problem solving polynomial equations

Solving Polynomials

"Solving" means finding the "roots" ...

... a "root" (or "zero") is where the function is equal to zero :

In between the roots the function is either entirely above, or entirely below, the x-axis

Example: −2 and 2 are the roots of the function x 2 − 4

Let's check:

• when x = −2, then x 2 − 4 = (−2) 2 − 4 = 4 − 4 = 0
• when x = 2, then x 2 − 4 = 2 2 − 4 = 4 − 4 = 0

How do we solve polynomials? That depends on the Degree !

The first step in solving a polynomial is to find its degree.

The Degree of a Polynomial with one variable is ...

... the largest exponent of that variable.

When we know the degree we can also give the polynomial a name:

How To Solve

So now we know the degree, how to solve?

• Read how to solve Linear Polynomials (Degree 1) using simple algebra.
• Read how to solve Quadratic Polynomials (Degree 2) with a little work,
• It can be hard to solve Cubic (degree 3) and Quartic (degree 4) equations,
• And beyond that it can be impossible to solve polynomials directly.

So what do we do with ones we can't solve? Try to solve them a piece at a time!

If we find one root, we can then reduce the polynomial by one degree (example later) and this may be enough to solve the whole polynomial.

Here are some main ways to find roots.

1. Basic Algebra

We may be able to solve using basic algebra:

Example: 2x+1

2x+1 is a linear polynomial:

The graph of y = 2x+1 is a straight line

It is linear so there is one root.

Use Algebra to solve:

A "root" is when y is zero: 2x+1 = 0

Subtract 1 from both sides: 2x = −1

Divide both sides by 2: x = −1/2

And that is the solution:

(You can also see this on the graph)

We can also solve Quadratic Polynomials using basic algebra (read that page for an explanation).

2. By experience, or simply guesswork.

It is always a good idea to see if we can do simple factoring:

Example: x 3 +2x 2 −x

This is cubic ... but wait ... we can factor out "x":

x 3 +2x 2 −x = x(x 2 +2x−1)

Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.

Example: x 3 −8

Again this is cubic ... but it is also the " difference of two cubes ":

x 3 −8 = x 3 −2 3

And so we can turn it into this:

x 3 −8 = (x−2)(x 2 +2x+4)

There is a root at x=2, because:

(2−2)(2 2 +2×2+4) = (0) (2 2 +2×2+4)

And we can then solve the quadratic x 2 +2x+4 and we are done

3. Graphically.

Graph the polynomial and see where it crosses the x-axis.

Graphing is a good way to find approximate answers, and we may also get lucky and discover an exact answer.

Caution: before you jump in and graph it, you should really know How Polynomials Behave , so you find all the possible answers!

This is useful to know: When a polynomial is factored like this:

f(x) = (x−a)(x−b)(x−c)...

Then a, b, c, etc are the roots !

So Linear Factors and Roots are related, know one and we can find the other.

(Read The Factor Theorem for more details.)

Example: f(x) = (x 3 +2x 2 )(x−3)

We see "(x−3)", and that means that 3 is a root (or "zero") of the function.

Well, let us put "3" in place of x:

f(x) = (3 3 +2·3 2 )(3−3)

f(3) = (3 3 +2·3 2 )( 0 )

Yes! f(3)=0, so 3 is a root.

How to Check

Found a root? Check it!

Simply put the root in place of "x": the polynomial should be equal to zero.

Example: 2x 3 −x 2 −7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2 . We can check easily, just put "2" in place of "x":

f(2) = 2(2) 3 −(2) 2 −7(2)+2 = 16−4−14+2 = 0

Yes! f(2)=0 , so we have found a root!

How about where it crosses near −1.8 :

f(−1.8) = 2(−1.8) 3 −(−1.8) 2 −7(−1.8)+2 = −11.664−3.24+12.6+2 = −0.304

No, it isn't equal to zero, so −1.8 will not be a root (but it may be close!)

But we did discover one root, and we can use that to simplify the polynomial, like this

Example (continued): 2x 3 −x 2 −7x+2

So, f(2)=0 is a root ... that means we also know a factor:

(x−2) must be a factor of 2x 3 −x 2 −7x+2

Next, divide 2x 3 −x 2 −7x+2 by (x−2) using Polynomial Long Division to find:

2x 3 −x 2 −7x+2 = (x−2)(2x 2 +3x−1)

So now we can solve 2x 2 +3x−1 as a Quadratic Equation and we will know all the roots.

That last example showed how useful it is to find just one root. Remember:

If we find one root, we can then reduce the polynomial by one degree and this may be enough to solve the whole polynomial.

How Far Left or Right

When trying to find roots, how far left and right of zero should we go?

There is a way to tell, and there are a few calculations to do, but it is all simple arithmetic. Read Bounds on Zeros for all the details.

Have We Got All The Roots?

There is an easy way to know how many roots there are. The Fundamental Theorem of Algebra says:

A polynomial of degree n ... ... has n roots (zeros) but we may need to use complex numbers

So: number of roots = the degree of polynomial .

Example: 2x 3 + 3x − 6

The degree is 3 (because the largest exponent is 3), and so:

There are 3 roots.

But Some Roots May Be Complex

Yes, indeed, some roots may be complex numbers (ie have an imaginary part), and so will not show up as a simple "crossing of the x-axis" on a graph.

But there is an interesting fact:

Complex Roots always come in pairs !

So we either get no complex roots, or 2 complex roots, or 4 , etc... Never an odd number.

Which means we automatically know this:

Positive or Negative Roots?

There is also a special way to tell how many of the roots are negative or positive called the Rule of Signs that you may like to read about.

Multiplicity of a Root

Sometimes a factor appears more than once. We call that Multiplicity :

• Multiplicity is how often a certain root is part of the factoring.

Example: f(x) = (x−5) 3 (x+7)(x−1) 2

This could be written out in a more lengthy way like this:

f(x) = (x−5)(x−5)(x−5)(x+7)(x−1)(x−1)

(x−5) is used 3 times, so the root "5" has a multiplicity of 3 , likewise (x+7) appears once and (x−1) appears twice. So:

• the root +5 has a multiplicity of 3
• the root −7 has a multiplicity of 1 (a "simple" root)
• the root +1 has a multiplicity of 2

Q: Why is this useful? A: It makes the graph behave in a special way!

When we see a factor like (x-r) n , "n" is the multiplicity, and

• even multiplicity just touches the axis at "r" (and otherwise stays one side of the x-axis)
• odd multiplicity crosses the axis at "r" (changes from one side of the x-axis to the other)

We can see it on this graph:

Number Line

• -4x^3+6x^2+2x=0
• 6+11x+6x^2+x^3=0
• 2x^5+x^4-2x-1=0
• 11+6x+x^2=-\frac{6}{x}
• How do you solve polynomials equations?
• To solve a polynomial equation write it in standard form (variables and canstants on one side and zero on the other side of the equation). Factor it and set each factor to zero. Solve each factor. The solutions are the solutions of the polynomial equation.
• What is polynomial equation?
• A polynomial equation is an equation formed with variables, exponents and coefficients. The highest exponent is the order of the equation.
• What is not polynomial?
• A non-polynomial function or expression is one that cannot be written as a polynomial. Non-polynomial functions include trigonometric functions, exponential functions, logarithmic functions, root functions, and more.
• Can 0 be a polynomial?
• Like any constant zero can be considered as a constant polynimial. It is called the zero polynomial and have no degree.

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• High School Math Solutions – Exponential Equation Calculator Solving exponential equations is pretty straightforward; there are basically two techniques: <ul> If the exponents...

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How to Solve Polynomial Equations

Copyright © 2002–2024 by Stan Brown, BrownMath.com

Summary: In algebra you spend lots of time solving polynomial equations or factoring polynomials (which is the same thing). It would be easy to get lost in all the techniques, but this paper ties them all together in a coherent whole.

Factor = Root

Exact or approximate, step by step, cubic and quartic formulas, step 1. standard form and simplify, descartes’ rule of signs, complex roots, irrational roots, multiple roots, step 3. quadratic factors, monomial factors, special products, rational roots, graphical clues, boundaries on roots, step 5. divide by your factor, web calculators, ti calculators, complete example, what’s new, the master plan.

Make sure you aren’t confused by the terminology. All of these are the same:

• Solving a polynomial equation p ( x ) = 0
• Finding roots of a polynomial equation p ( x ) = 0
• Finding zeroes of a polynomial function p ( x )
• Factoring a polynomial function p ( x )

There’s a factor for every root, and vice versa. ( x − r ) is a factor if and only if r is a root. This is the Factor Theorem : finding the roots or finding the factors is essentially the same thing. (The main difference is how you treat a constant factor .)

Most often when we talk about solving an equation or factoring a polynomial, we mean an exact (or analytic) solution . The other type, approximate (or numeric) solution , is always possible and sometimes is the only possibility.

When you can find it, an exact solution is usually better . You can always find a numerical approximation to an exact solution, but going the other way is much more difficult. This page spends most of its time on methods for exact solutions, but also tells you what to do when analytic methods fail — or when you actually want an approximate solution, as in many engineering and science problems .

How do you find the factors or zeroes of a polynomial (or the roots of a polynomial equation)? Basically, you whittle . Every time you chip a factor or root off the polynomial, you’re left with a polynomial that is one degree simpler. Use that new reduced polynomial to find the remaining factors or roots.

Follow this procedure step by step:

If you’re down to a cubic or quartic equation (degree 3 or 4), you have a choice of continuing with factoring (step 4) or using the cubic or quartic formulas . These formulas are a lot of work, so most people prefer to keep factoring.

• Find one rational factor or root. This is the hard part, but there are lots of techniques to help you.   [  details  ] If you can find a factor or root, continue with step 5 below; if you can’t (or if you actually want an approximate solution) , go to step 6.
• Divide by your factor . This leaves you with a new reduced polynomial whose degree is 1 less.   [  details  ]   For the rest of the problem, you’ll work with the reduced polynomial and not the original. Continue at step 3.
• If you can’t find a factor or root , turn to numerical methods.   [  details  ] Then go to step 7.
• If this was an equation to solve, write down the roots . If it was a polynomial to factor, write it in factored form , including any constant factors you took out in step 1.

This is an example of an algorithm , a set of steps that will lead to a desired result in a finite number of operations. It’s an iterative strategy, because the middle steps are repeated as long as necessary.

The methods given here—find a rational root and use synthetic division—are the easiest. But if you can’t find a rational root, there are special methods for cubic equations (degree 3) and quartic equations (degree 4), both at Mathworld. An alternative approach is provided by Dick Nickalls in PDF for cubic and quartic equations.

This is an easy step—easy to overlook, unfortunately. If you have a polynomial equation , put all terms on one side and 0 on the other. And whether it’s a factoring problem or an equation to solve, put your polynomial in standard form, from highest to lowest power .

For instance, you cannot solve this equation in this form:

x ³ + 6 x ² + 12 x = −8

You must change it to this form:

x ³ + 6 x ² + 12 x + 8 = 0

Also make sure you have simplified, by factoring out any common factors . This may include factoring out a −1 so that the highest power has a positive coefficient. Example: to factor

7 − 6 x − 15 x ² − 2 x ³

begin by putting it in standard form:

−2 x ³ − 15 x ² − 6 x + 7

and then factor out the −1

−(2 x ³ + 15 x ² + 6 x − 7) or (−1)(2 x ³ + 15 x ² + 6 x − 7)

If you’re solving an equation, you can throw away any common constant factor. (Technically, you’re dividing left and right sides by that constant factor.) But if you’re factoring a polynomial, you must keep the common factor .

Example: To solve 8 x ² + 16 x  + 8 = 0, you can divide left and right by the common factor 8. The equation x ² + 2 x  + 1 = 0 has the same roots as the original equation .

But to factor 8 x ² + 16 x  + 8 , you recognize the common factor of 8 and rewrite the polynomial as 8( x ² + 2 x  + 1), which is identical to the original polynomial . (While you will focus your further factoring efforts on x ² + 2 x  + 1, it would be an error to write that the original polynomial equals x ² + 2 x  + 1.)

Your “common factor” may be a fraction, because you must factor out any fractions so that the polynomial has integer coefficients .

Example: To solve (1/3) x ³ + (3/4) x ² − (1/2) x  + 5/6 = 0, you recognize the common factor of 1/12 and divide both sides by 1/12. This is exactly the same as recognizing and multiplying by the lowest common denominator of 12. Either way, you get 4 x ³ + 9 x ² − 6 x  + 10 = 0, which has the same roots as the original equation .

Step 2. How Many Roots?

A polynomial of degree n will have n roots, some of which may be multiple roots .

How do you know this is true? The Fundamental Theorem of Algebra tells you that the polynomial has at least one root. The Factor Theorem tells you that if r is a root then ( x − r ) is a factor. But if you divide a polynomial of degree n by a factor ( x − r ), whose degree is 1, you get a polynomial of degree n −1. Repeatedly applying the Fundamental Theorem and Factor Theorem gives you n roots and n factors (not necessarily all different) .

Descartes’ Rule of Signs can tell you how many positive and how many negative real zeroes the polynomial has. This is a big labor-saving device, especially when you’re deciding which possible rational roots to pursue.

p ( x ) = x 5 − 2 x 3 + 2 x 2 − 3 x + 12

has four variations in sign.

Descartes’ Rule of Signs:

Example: Consider p ( x ) above. Since it has four variations in sign, there must be either four positive roots, two positive roots, or no positive roots.

p (− x ) = (− x ) 5  − 2(− x ) 3  + 2(− x ) 2  − 3(− x ) + 12

p (− x ) = − x 5  + 2 x 3  + 2 x 2  + 3 x  + 12

p (− x ) has one variation in sign, and therefore the original p ( x ) has one negative root. Since you know that p ( x ) must have a negative root, but it may or may not have any positive roots, you would look first for negative roots.

p ( x ) is a fifth-degree polynomial, and therefore it must have five zeros. Since x is not a factor, you know that x  = 0 is not a zero of the polynomial. (For a polynomial with real coefficients, like this one, complex roots occur in pairs .) Therefore there are three possibilities:

If a polynomial has real coefficients , then either all roots are real or there are an even number of non-real complex roots, in conjugate pairs .

For example, if 5+2i is a zero of a polynomial with real coefficients, then 5−2i must also be a zero of that polynomial. It is equally true that if ( x −5−2i) is a factor then ( x −5+2i) is also a factor.

Why is this true? Because when you have a factor with an imaginary part and multiply it by its complex conjugate you get a real result:

( x −5−2i)( x −5+2i) = x ²−10 x +25−4i² = x ²−10 x +29

If ( x −5−2i) was a factor but ( x −5+2i) was not, then the polynomial would end up with imaginaries in its coefficients, no matter what the other factors might be. If the polynomial has only real coefficients, then any complex roots must occur in conjugate pairs.

For similar reasons, if the polynomial has rational coefficients then the irrational roots involving square roots occur (if at all) in conjugate pairs. If ( x −2+√3) is a factor of a polynomial with rational coefficients, then ( x −2−√3) must also be a factor. To see why, remember how you rationalize a binomial denominator; or just check what happens when you multiply those two factors.

As Jeff Beckman pointed out (20 June 2006), this is emphatically not true for odd roots. For instance, x ³−2 = 0 has three roots: 3 √ 2 and two complex roots.

It’s an interesting problem whether irrationals involving even roots of order ≥4 must also occur in conjugate pairs. I don’t have an immediate answer.

When a given factor ( x − r ) occurs m times in a polynomial, r is called a multiple root or a root of multiplicity m .

Examples: Compare these two polynomials and their graphs:

f ( x ) =  ( x −1)( x −4) 2  = x 3  − 9 x 2  + 24 x  − 16

g ( x ) = ( x −1) 3 ( x −4) 2  = x 5 − 11 x 4 + 43 x 3 − 73 x 2 + 56 x − 16

These polynomials have the same zeroes, but the root 1 occurs with different multiplicities. Look at the graphs:

Both polynomials have zeroes at 1 and 4 only. f ( x ) has degree 3, which means three roots. You see from the factors that 1 is a root of multiplicity 1 and 4 is a root of multiplicity 2. Therefore the graph crosses the axis at x  = 1 (but is not horizontal there) and touches at x  = 4 without crossing.

By contrast, g ( x ) has degree 5. ( g ( x ) = f ( x ) times ( x −1) 2 .) Of the five roots, 1 occurs with multiplicity 3: the graph crosses the axis at x  = 1 and is horizontal there; 4 occurs with multiplicity 2, and the graph touches the axis at x  = 4 without crossing.

When you have quadratic factors (Ax²+Bx+C), it may or may not be possible to factor them further.

For example, suppose you have a factor of 12 x ²− x −35. Can that be factored further? By trial and error you’d have to try a lot of combinations! Instead, use the fact that factors correspond to roots , and apply the formula to find the roots of 12 x ²− x −35 = 0, like this:

x = [ −(−1) ± √ 1 − 4(12)(−35) ] / 2(12)

x = [ 1 ± √ 1681 ] / 24

√ 1681 = 41, and therefore

x = [ 1 ± 41 ] / 24

x = 42/24 or −40/24

x = 7/4 or −5/3

If 7/4 and −5/3 are roots, then ( x −7/4) and ( x +5/3) are factors. Therefore

12 x ²− x −35 = 12( x −7/4)( x +5/3) or (4 x −7)(3 x +5)

What about x ²−5 x +7? This one looks like it’s prime, but how can you be sure? Again, apply the formula:

x = [ −(−5) ± √ 25 − 4(1)(7) ] / 2(1)

x = [ 5 ± √ −3 ] / 2

What you do with that depends on the original problem. If it was to factor over the reals, then x ²−5 x +7 is prime. But if that factor was part of an equation and you were supposed to find all complex roots, you have two of them:

x = 5/2 + (√ 3 /2)i, x = 5/2 − (√ 3 /2)i

Since the original equation had real coefficients, these complex roots occur in a conjugate pair .

Step 4. Find One Factor or Root

This step is the heart of factoring a polynomial or solving a polynomial equation. There are a lot of techniques that can help you to find a factor.

Sometimes you can find factors by inspection (see the first two sections that follow). This provides a great shortcut, so check for easy factors before starting more strenuous methods .

f ( x ) = 4 x 6  + 12 x 5  + 12 x 4  + 4 x 3

you should immediately factor it as

f ( x ) = 4 x 3 ( x 3  + 3 x 2  + 3 x  + 1)

Getting the 4 out of there simplifies the remaining numbers, the x 3 gives you a root of x  = 0 (with multiplicity 3), and now you have only a cubic polynomial (degree 3) instead of a sextic (degree 6). In fact, you should now recognize that cubic as a special product , the perfect cube ( x +1) 3 .

When you factor out a common variable factor, be sure you remember it at the end when you’re listing the factor or roots. x ³+3 x ²+3 x +1 = 0 has certain roots, but x ³( x ³+3 x ²+3 x +1) = 0 has those same roots and also a root at x  = 0 (with multiplicity 3) .

Be alert for applications of the Special Products . If you can apply them, your task becomes much easier. The Special Products are

• perfect square (2 forms): A ² ± 2 A B + B ² = ( A ± B )²
• sum of squares: A ² + B ² cannot be factored on the reals, in general (for exceptional cases see How to Factor the Sum of Squares )
• difference of squares: A ² − B ² = ( A + B )( A − B )
• perfect cube (2 forms): A ³ ± 3 A ² B + 3 A B ² ± B ³ = ( A ± B )³
• sum of cubes: A ³ + B ³ = ( A + B )( A ² − A B + B ²)
• difference of cubes: A ³ − B ³ = ( A − B )( A ² + A B + B ²)

The expressions for the sum or difference of two cubes look as though they ought to factor further, but they don’t. A ²± A B + B ² is prime over the reals.

p ( x ) = 27 x ³ − 64

You should recognize this as

p ( x ) = (3 x )³ − 4³

You know how to factor the difference of two cubes:

p ( x ) = (3 x −4)(9 x ²+12 x +16)

Bingo! As soon as you get down to a quadratic, you can apply the Quadratic Formula and you’re done.

Here’s another example:

q ( x ) = x 6  + 16 x 3  + 64

This is just a perfect square trinomial, but in x 3 instead of x . You factor it exactly the same way:

q ( x ) = ( x 3 ) 2 + 2(8)( x 3 ) + 8 2

q ( x ) = ( x 3  + 8) 2

And you can easily factor ( x 3 +8) 2 as ( x +2) 2 ( x 2 −2 x +4) 2 .

Assuming you’ve already factored out the easy monomial factors and special products , what do you do if you’ve still got a polynomial of degree 3 or higher?

The answer is the Rational Root Test . It can show you some candidate roots when you don’t see how to factor the polynomial, as follows.

f ( x ) = a n x n + … + a o

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction p / q , where p is a factor of the trailing constant a o and q is a factor of the leading coefficient a n .

f ( x ) = 2 x 4 − 11 x 3 − 6 x 2 + 64 x + 32

The factors of the leading coefficient (2) are 2 and 1. The factors of the constant term (32) are 1, 2, 4, 8, 16, and 32. Therefore the possible rational zeroes are ±1, 2, 4, 8, 16, or 32 divided by 2 or 1:

± any of 1/2, 1/1, 2/2, 2/1, 4/2, 4/1, 8/2, 8/1, 16/2, 16/1, 32/2, 32/1

reduced: ± any of ½, 1, 2, 4, 8, 16, 32

What do we mean by saying this is a list of all the possible rational roots ? We mean that no other rational number, like ¼ or 32/7, can be a zero of this particular polynomial.

Caution : Don’t make the Rational Root Test out to be more than it is. It doesn’t say those rational numbers are roots, just that no other rational numbers can be roots. And it doesn’t tell you anything about whether some irrational or even complex roots exist. The Rational Root Test is only a starting point.

Suppose you have a polynomial with non-integer coefficients. Are you stuck? No, you can factor out the least common denominator (LCD) and get a polynomial with integer coefficients that way. Example:

(1/2) x ³ − (3/2) x ² + (2/3) x − 1/2

The LCD is 1/6. Factoring out 1/6 gives the polynomial

(1/6)(3 x ³ − 9 x ² + 4 x − 3)

The two forms are equivalent, and therefore they have the same roots. But you can’t apply the Rational Root Test to the first form, only to the second. The test tells you that the only possible rational roots are ± any of 1/3, 1, 3.

Once you’ve identified the possible rational zeroes, how can you screen them? The brute-force method would be to take each possible value and substitute it for x in the polynomial: if the result is zero then that number is a root. But there’s a better way.

Use synthetic division to see if each candidate makes the polynomial equal zero. This is better for three reasons. First, it’s computationally easier, because you don’t have to compute higher powers of numbers. Second, at the same time it tells you whether a given number is a root, it produces the reduced polynomial that you’ll use to find the remaining roots. Finally, the results of synthetic division may give you an upper or lower bound even if the number you’re testing turns out not to be a root.

Sometimes Descartes’ Rule of Signs can help you screen the possible rational roots further. For example, the Rational Root Test tells you that if

g ( x ) = 2 x 4 + 13 x 3 + 20 x 2 + 28 x + 8

has any rational roots, they must come from the list ± any of ½, 1, 2, 4, 8. But don’t just start off substituting or synthetic dividing. Since there are no sign changes, there are no positive roots. Are there any negative roots?

g (− x ) = 2 x 4 − 13 x 3 + 20 x 2 − 28 x + 8

has four sign changes. Therefore there could be as many as four negative roots. (There could also be two negative roots, or none.) There’s no guarantee that any of the roots are rational, but any root that is rational must come from the list −½, −1, −2, −4, −8.

(If you have a graphing calculator, you can pre-screen the rational roots by graphing the polynomial and seeing where it seems to cross the x  axis. But you still need to verify the root algebraically, to see that g ( x ) is exactly 0 there, not just nearly 0.)

Remember, the Rational Root Test guarantees to find all rational roots. But it will completely miss real roots that are not rational, like the roots of x ² − 2 = 0, which are ±√2, or the roots of x ² + 4 = 0, which are ±2i.

p ( x ) = 2 x 4 − 11 x 3 − 6 x 2 + 64 x + 32

The Rational Root Theorem tells you that the only possible rational zeroes are ±½, 1, 2, 4, 8, 16, 32. But suppose you factor out the 2 (as I once did in class), writing the equivalent function

p ( x ) = 2( x 4 − (11/2) x 3 − 3 x 2 + 32 x + 16)

This function is the same as the earlier one, but you can no longer apply the Rational Root Test because the coefficients are not integers. In fact −½ is a zero of p ( x ), but it did not show up when I (illegally) applied the Rational Root Test to the second form. My mistake was forgetting that the Rational Root Theorem applies only when all coefficients of the polynomial are integers.

By graphing the function—either by hand or with a graphing calculator—you can get a sense of where the roots are, approximately, and how many real roots exist.

Example: If the Rational Root Test tells you that ±2 are possible rational roots, you can look at the graph to see if it crosses or touches the x  axis at 2 or −2. If so, use synthetic division to verify that the suspected root actually is a root. Yes, you always need to check—from the graph you can never be sure whether the intercept is at your possible rational root or just near it.

Some techniques don’t tell you the specific value of a root, but rather that a root exists between two values or that all roots are less than a certain number of greater than a certain number. This helps narrow down your search.

Intermediate Value Theorem

This theorem tells you that if the graph of a polynomial is above the x  axis for one value of x and below the x  axis for another value of x , it must cross the x  axis somewhere between. (If you can graph the function , the crossings will usually be obvious.)

p ( x ) = 3 x ³ + 4 x ² − 20 x −32

The rational roots (if any) must come from the list ± any of 1/3, 2/3, 1, 4/3, 2, 8/3, 4, 16/3, 8, 32/3, 16, 32. Naturally you’ll look at the integers first, because the arithmetic is easier. Trying synthetic division , you find p (1) = −45, p (2) = −22, and p (4) = 144. Since p (2) and p (4) have opposite signs, you know that the graph crosses the axis between x  = 2 and x  = 4, so there is at least one root between those numbers. In other words, either 8/3 is a root, or the root(s) between 2 and 4 are irrational. (In fact, synthetic division reveals that 8/3 is a root.)

The Intermediate Value Theorem can tell you where there is a root, but it can’t tell you where there is no root. For example, consider

q ( x ) = 4 x ² − 16 x + 15

q (1) and q (3) are both positive, but that doesn’t tell you whether the graph might touch or cross the axis between. (It actually crosses the axis twice, at x  = 3/2 and x  = 5/2.)

Upper and Lower Bounds

One side effect of synthetic division is that even if the number you’re testing turns out not to be a root, it may tell you that all the roots are smaller or larger than that number:

• If you do synthetic division by a positive number a , and every number in the bottom row is positive or zero, then a is an upper bound for the roots, meaning that all the real roots are ≤  a .

What if the bottom row contains zeroes? A more complete statement is that alternating nonnegative and nonpositive signs , after synthetic division by a negative number, show a lower bound on the root. The next two examples clarify that.

(By the way, the rule for lower bounds follows from the rule for upper bounds. Lower limits on roots of p ( x ) equal upper limits on roots of p (− x ), and dividing by (− x + r ) is the same as dividing by −( x − r ).)

q ( x ) = x 3  + 2 x 2  − 3 x  − 4

Using the Rational Root Test , you identify the only possible rational roots as ±4, ±2, and ±1. You decide to try −2 as a possible root, and you test it with synthetic division:

−2 is not a root of the equation f ( x ) = 0. The third row shows alternating signs, and you were dividing by a negative number; however, that zero mucks things up. Recall that you have a lower bound only if the signs in the bottom row alternate nonpositive and nonnegative. The 1 is positive (nonnegative), and the 0 can count as nonpositive, but the −3 doesn’t qualify as nonnegative. The alternation is broken, and you do not know whether there are roots smaller than −2. (In fact, graphical or numerical methods would show a root around −2.5.) Therefore you need to try the lower possible rational root, −4:

Here the signs do alternate; therefore you know there are no roots below −4. (The remainder −24 shows you that −4 itself isn’t a root.)

r ( x ) = x ³ + 3 x ² − 3

The Rational Root Test tells you that the possible rational roots are ±1 and ±3. With synthetic division for −3:

−3 is not a root, but the signs do alternate here, since the first 0 counts as nonpositive and the second as nonnegative. Therefore −3 is a lower bound to the roots, meaning that the equation has no real roots lower than −3.

Coefficients and Roots

There is an interesting relationship between the coefficients of a polynomial and its zeroes. I mention it last because it is more suited to forming a polynomial that has zeroes with desired properties, rather than finding zeroes of an existing polynomial. However, if you know all the roots of a polynomial but one or two, you can easily use this technique to find the remaining root.

Consider the polynomial

f ( x ) = a n x n + a n −1 x n −1 + a n −2 x n −2 + … + a 2 x 2 + a 1 x + a o

The following relationships exist:

Example: f ( x ) = x 3  − 6 x 2  − 7 x  − 8 has degree 3, and therefore at most three real zeroes. If we write the real zeroes as r 1 , r 2 , r 3 , then the sum of the roots is r 1  + r 2  + r 3  = −(−6) = 6; the sum of the products of roots taken two at a time is r 1 r 2  + r 1 r 3  + r 2 r 3  = −7, and the product of the roots is r 1 r 2 r 3  = (−1) 3 (−8) = 8.

Example: Given that the polynomial

g ( x ) = x 5 − 11 x 4 + 43 x 3 − 73 x 2 + 56 x − 16

has a triple root at x  = 1, find the other two roots.

Solution: Let the other two roots be c and d . Then you know that the sum of the all roots is 1 + 1 + 1 + c  + d  = −(−11) = 11, or c  + d  = 8. You also know that the product of all the roots is 1 × 1 × 1 × c d  = (−1) 5 (−16) = 16, or c d  = 16. c  + d  = 8, c d  = 16; therefore c  = d  = 4, so the remaining roots are a double root at x  = 4.

More Coefficients and Roots

There are several further theorems about the relationship between coefficients and roots. Wikipedia’s article Properties of Polynomial Roots gives a good though somewhat terse summary.

Remember that r is a root if and only if x − r is a factor; this is the Factor Theorem . So if you want to check whether r is a root, you can divide the polynomial by x − r and see whether it comes out even (remainder of 0). Elizabeth Stapel has a nice example of dividing polynomials by long division.

But it’s easier and faster to do synthetic division. If your synthetic division is a little rusty, you might want to look at Dr. Math’s short Synthetic Division tutorial ; if you need a longer tutorial, Elizabeth Stapel’s Synthetic Division is excellent. (Dr. Math also has a page on why Synthetic Division works .)

Synthetic division also has some side benefits. If your suspected root actually is a root, synthetic division gives you the reduced polynomial . And sometimes you also luck out and synthetic division shows you an upper or lower bound on the roots.

You can use synthetic division when you’re dividing by a binomial of the form x − r for a constant r . If you’re dividing by x −3, you’re testing whether 3 is a root and you synthetic divide by 3 (not −3). If you’re dividing by x +11, you’re testing whether −11 is a root and you synthetic divide by −11 (not 11).

p ( x ) = 4 x 4  − 35 x 2  − 9

You suspect that x −3 might be a factor, and you test it by synthetic division, like this:

Since the remainder is 0, you know that 3 is a root of p ( x ) = 0, and x −3 is a factor of p ( x ). But you know more. Since 3 is positive and the bottom row of the synthetic division is all positive or zero, you know that all the roots of p ( x ) = 0 must be ≤ 3. And you also know that

p ( x ) = ( x −3)(4 x 3  + 12 x 2  +  x  + 3)

4 x 3  + 12 x 2  +  x  + 3 is the reduced polynomial . All of its factors are also factors of the original p ( x ), but its degree is one lower , so it’s easier to work with.

Step 6. Numerical Methods

When your equation has no more rational roots (or your polynomial has no more rational factors), you can turn to numerical methods to find the approximate value of irrational roots:

• Newton’s method converges quickly, but the derivative must exist and be continuous, and of course you need to know how to find the derivative. Wikipedia has a decent article on the method, and my article Newton’s Method on TI-83/84 or TI-89 walks you through the calculator procedure.
• The Regula falsi method can be slower than Newton’s method, but it doesn’t have the limitations of Newton’s method.

A search like this one will find a bazillion online polynomial calculators. However, many of them fail in one way or another on the example below : they miss the complex roots, or they can’t show the steps in the calculation, or ask for money to show the steps.

MathPortal’s Polynomial Equation Solver is an excellent free resource, and it did a fine job with that example .

If you have a TI-83 or TI-84 , you can get the zeroes of a polynomial numerically. Graph the polyomial, then use calc  » zero to find the real zeroes. This YouTube video shows you the process. (This won’t help you with the complex ones, if any.)

Recent versions of the TI-84 , beginning with the TI-84 Silver Edition, have APPS  » PlySmlt2  » POLYNOMIAL ROOT FINDER , which includes an option to show complex roots. This YouTube video shows the process and gives some tips for the black&white calculator, and this one does the same for the color TI-84s.

The TI-89 will give you exact solutions, if possible, for real and complex roots. (You may need to press the MODE key, scroll down to Exact/Approx , and change it to Exact.)

Select F2:Algebra  » A:Complex  » cSolve , then enter your equation including = 0, press the comma key, and then the name of your variable, and finally press ) and ENTER. For the example below , the input line should look like this:

cSolve(4x³+15x−36=0,x)

This video shows how to get exact solutions or numeric solutions on the TI-89.

Solve for all complex roots:

4 x ³ + 15 x − 36 = 0

(We’ll call the left-hand side f ( x ).)

Step 1.   The equation is already in standard form, with only zero on one side, and powers of x from highest to lowest. There are no common factors.

Step 2.   Since the equation has degree 3, there will be 3 roots. There is one variation in sign, and from Descartes’ Rule of Signs you know there must be one positive root. Examine the polynomial with − x replacing x :

f (− x ) = −4 x ³ − 15 x − 36

There are no variations in sign, which means there are no negative roots. The other two roots must therefore be complex, and conjugates of each other.

Steps 3 and 4.   The possible rational roots are unfortunately rather numerous: any of 1, 2, 3, 4, 6, 9, 12, 18, 36 divided by any of 4, 2, 1. (Only positive roots are listed because you have already determined that there are no negative roots for this equation.) You decide to try 1 first:

1 is not a root, so you test 2:

Alas, 2 is not a root either. But notice that f (1) = −17 and f (2) = 26. They have opposite signs, which means that the graph crosses the x  axis between x  = 1 and x  = 2, and a root is between 1 and 2. (In this case it’s the only root, since you have determined that there is one positive root and there are no negative roots.)

The only possible rational root between 1 and 2 is 3/2, and therefore either 3/2 is a root or the root is irrational. You try 3/2 by synthetic division:

Hooray! 3/2 is a root. The reduced polynomial is 4 x ² + 6 x  + 24. In other words,

(4 x ³ + 15 x  − 36) ÷ ( x −3/2) = 4 x ² + 6 x  + 24

The reduced polynomial has degree 2, so there is no need for more trial and error, and you continue to step 5.

Step 5.   Now you must solve

4 x ² + 6 x  + 24 = 0

First divide out the common factor of 2:

2 x ² + 3 x  + 12 = 0

It’s no use trying to factor that quadratic, because you determined using Descartes’ Rule of Signs that there are no more real roots. So you use the quadratic formula :

x = [ −3 ± √ 9 − 4(2)(12) ] / 2(2)

x = [ −3 ± √ −87 ] / 4

x = −3/4 ± (√ 87 /4)i

Step 6.   Remember that you found a root in an earlier step! The full list of roots is

3/2,  −3/4 + (√ 87 /4)i,  −3/4 − (√ 87 /4)i

• 26 Mar 2022 : Rewrote step 6 to recommend Newton’s method and the Regula Falsi method specifically, and added a Web calculator and TI calculator methods . Added “usually” to the statement that exact solutions are better, and suggested that numeric solutions may be appropriate for engineering and science Made a number of text changes for clarity.
• 14/15 Nov 2021 : Updated links here and here . Updated all http: links to https:.
• (intervening changes suppressed)
• 15 Feb 2002 : First publication.

Updates and new info: https://BrownMath.com/alge/

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6.5 Polynomial Equations

Learning objectives.

By the end of this section, you will be able to:

Use the Zero Product Property

• Solve quadratic equations by factoring
• Solve equations with polynomial functions
• Solve applications modeled by polynomial equations

Be Prepared 6.10

Before you get started, take this readiness quiz.

Solve: 5 y − 3 = 0 . 5 y − 3 = 0 . If you missed this problem, review Example 2.2 .

Be Prepared 6.11

Factor completely: n 3 − 9 n 2 − 22 n . n 3 − 9 n 2 − 22 n . If you missed this problem, review Example 3.48 .

Be Prepared 6.12

If f ( x ) = 8 x − 16 , f ( x ) = 8 x − 16 , find f ( 3 ) f ( 3 ) and solve f ( x ) = 0 . f ( x ) = 0 . If you missed this problem, review Example 3.59 .

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form a x + b = c . a x + b = c .

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get n 2 + n . n 2 + n .

The general form of a quadratic equation is a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , with a ≠ 0 . a ≠ 0 . (If a = 0 , a = 0 , then 0 · x 2 = 0 0 · x 2 = 0 and we are left with no quadratic term.)

Quadratic Equation

An equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 is called a quadratic equation.

To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

We will first solve some quadratic equations by using the Zero Product Property . The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Product Property

If a · b = 0 , a · b = 0 , then either a = 0 a = 0 or b = 0 b = 0 or both.

We will now use the Zero Product Property, to solve a quadratic equation .

Example 6.44

How to solve a quadratic equation using the zero product property.

Solve: ( 5 n − 2 ) ( 6 n − 1 ) = 0 . ( 5 n − 2 ) ( 6 n − 1 ) = 0 .

Try It 6.87

Solve: ( 3 m − 2 ) ( 2 m + 1 ) = 0 . ( 3 m − 2 ) ( 2 m + 1 ) = 0 .

Try It 6.88

Solve: ( 4 p + 3 ) ( 4 p − 3 ) = 0 . ( 4 p + 3 ) ( 4 p − 3 ) = 0 .

Use the Zero Product Property.

• Step 1. Set each factor equal to zero.
• Step 2. Solve the linear equations.
• Step 3. Check.

Solve Quadratic Equations by Factoring

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we must be sure to start with the quadratic equation in standard form , a x 2 + b x + c = 0 . a x 2 + b x + c = 0 . Then we must factor the expression on the left.

Example 6.45

How to solve a quadratic equation by factoring.

Solve: 2 y 2 = 13 y + 45 . 2 y 2 = 13 y + 45 .

Try It 6.89

Solve: 3 c 2 = 10 c − 8 . 3 c 2 = 10 c − 8 .

Try It 6.90

Solve: 2 d 2 − 5 d = 3 . 2 d 2 − 5 d = 3 .

Solve a quadratic equation by factoring.

• Step 1. Write the quadratic equation in standard form, a x 2 + b x + c = 0 . a x 2 + b x + c = 0 .
• Step 2. Factor the quadratic expression.
• Step 3. Use the Zero Product Property.
• Step 4. Solve the linear equations.
• Step 5. Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form .

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Example 6.46

Solve: 169 x 2 = 49 . 169 x 2 = 49 .

We leave the check up to you.

Try It 6.91

Solve: 25 p 2 = 49 . 25 p 2 = 49 .

Try It 6.92

Solve: 36 x 2 = 121 . 36 x 2 = 121 .

In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property , one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Example 6.47

Solve: ( 3 x − 8 ) ( x − 1 ) = 3 x . ( 3 x − 8 ) ( x − 1 ) = 3 x .

Try It 6.93

Solve: ( 2 m + 1 ) ( m + 3 ) = 1 2 m . ( 2 m + 1 ) ( m + 3 ) = 1 2 m .

Try It 6.94

Solve: ( k + 1 ) ( k − 1 ) = 8 . ( k + 1 ) ( k − 1 ) = 8 .

In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Example 6.48

Solve: 3 x 2 = 12 x + 63 . 3 x 2 = 12 x + 63 .

Try It 6.95

Solve: 18 a 2 − 30 = −33 a . 18 a 2 − 30 = −33 a .

Try It 6.96

Solve: 123 b = −6 − 60 b 2 . 123 b = −6 − 60 b 2 .

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.

Example 6.49

Solve: 9 m 3 + 100 m = 60 m 2 . 9 m 3 + 100 m = 60 m 2 .

Try It 6.97

Solve: 8 x 3 = 24 x 2 − 18 x . 8 x 3 = 24 x 2 − 18 x .

Try It 6.98

Solve: 16 y 2 = 32 y 3 + 2 y . 16 y 2 = 32 y 3 + 2 y .

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

Example 6.50

For the function f ( x ) = x 2 + 2 x − 2 , f ( x ) = x 2 + 2 x − 2 ,

ⓐ find x when f ( x ) = 6 f ( x ) = 6 ⓑ find two points that lie on the graph of the function.

ⓑ Since f ( −4 ) = 6 f ( −4 ) = 6 and f ( 2 ) = 6 , f ( 2 ) = 6 , the points ( −4 , 6 ) ( −4 , 6 ) and ( 2 , 6 ) ( 2 , 6 ) lie on the graph of the function.

Try It 6.99

For the function f ( x ) = x 2 − 2 x − 8 , f ( x ) = x 2 − 2 x − 8 ,

ⓐ find x when f ( x ) = 7 f ( x ) = 7 ⓑ Find two points that lie on the graph of the function.

Try It 6.100

For the function f ( x ) = x 2 − 8 x + 3 , f ( x ) = x 2 − 8 x + 3 ,

ⓐ find x when f ( x ) = −4 f ( x ) = −4 ⓑ Find two points that lie on the graph of the function.

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

Zero of a Function

For any function f , if f ( x ) = 0 , f ( x ) = 0 , then x is a zero of the function .

When f ( x ) = 0 , f ( x ) = 0 , the point ( x , 0 ) ( x , 0 ) is a point on the graph. This point is an x -intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.

Example 6.51

For the function f ( x ) = 3 x 2 + 10 x − 8 , f ( x ) = 3 x 2 + 10 x − 8 , find

ⓐ the zeros of the function, ⓑ any x -intercepts of the graph of the function, ⓒ any y -intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0.

ⓑ An x -intercept occurs when y = 0 . y = 0 . Since f ( −4 ) = 0 f ( −4 ) = 0 and f ( 2 3 ) = 0 , f ( 2 3 ) = 0 , the points ( −4 , 0 ) ( −4 , 0 ) and ( 2 3 , 0 ) ( 2 3 , 0 ) lie on the graph. These points are x -intercepts of the function. ⓒ A y -intercept occurs when x = 0 . x = 0 . To find the y -intercepts we need to find f ( 0 ) . f ( 0 ) .

Since f ( 0 ) = −8 , f ( 0 ) = −8 , the point ( 0 , −8 ) ( 0 , −8 ) lies on the graph. This point is the y -intercept of the function.

Try It 6.101

For the function f ( x ) = 2 x 2 − 7 x + 5 , f ( x ) = 2 x 2 − 7 x + 5 , find

ⓐ the zeros of the function, ⓑ any x -intercepts of the graph of the function, ⓒ any y -intercepts of the graph of the function.

Try It 6.102

For the function f ( x ) = 6 x 2 + 13 x − 15 , f ( x ) = 6 x 2 + 13 x − 15 , find

Solve Applications Modeled by Polynomial Equations

The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

Use a problem solving strategy to solve word problems.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose a variable to represent that quantity.
• Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Step 5. Solve the equation using appropriate algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a polynomial equation.

Example 6.52

The product of two consecutive odd integers is 323. Find the integers.

Try It 6.103

The product of two consecutive odd integers is 255. Find the integers.

Try It 6.104

The product of two consecutive odd integers is 483 Find the integers.

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

Example 6.53

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

Try It 6.105

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

Try It 6.106

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

In the next example, we will use the Pythagorean Theorem ( a 2 + b 2 = c 2 ) . ( a 2 + b 2 = c 2 ) . This formula gives the relation between the legs and the hypotenuse of a right triangle.

We will use this formula to in the next example.

Example 6.54

A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.

Try It 6.107

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

Try It 6.108

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

Example 6.55

Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function h ( t ) = −16 t 2 + 64 t + 80 h ( t ) = −16 t 2 + 64 t + 80 models the height, h , of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the ball hits the ground, ⓑ when the ball will be 80 feet above the ground, ⓒ the height of the ball at t = 2 t = 2 seconds.

ⓐ The zeros of this function are found by solving h ( t ) = 0 . h ( t ) = 0 . This will tell us when the ball will hit the ground.

The result t = 5 t = 5 tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result t = −1 t = −1 is discarded.

ⓑ The ball will be 80 feet above the ground when h ( t ) = 80 . h ( t ) = 80 .

ⓒ To find the height ball at t = 2 t = 2 seconds we find h ( 2 ) . h ( 2 ) .

Try It 6.109

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function h ( t ) = −16 t 2 + 48 t + 160 h ( t ) = −16 t 2 + 48 t + 160 models the height, h , of the rock above the ocean as a function of time, t . Find:

ⓐ the zeros of this function which tell us when the rock will hit the ocean, ⓑ when the rock will be 160 feet above the ocean, ⓒ the height of the rock at t = 1.5 t = 1.5 seconds.

Try It 6.110

Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 128 h ( t ) = −16 t 2 + 32 t + 128 models the height, h , of the penny above the ocean as a function of time, t . Find:

ⓐ the zeros of this function which is when the penny will hit the ocean, ⓑ when the penny will be 128 feet above the ocean, ⓒ the height the penny will be at t = 1 t = 1 seconds which is when the penny will be at its highest point.

Access this online resource for additional instruction and practice with quadratic equations.

• Beginning Algebra & Solving Quadratics with the Zero Property

Section 6.5 Exercises

Practice makes perfect.

In the following exercises, solve.

( 3 a − 10 ) ( 2 a − 7 ) = 0 ( 3 a − 10 ) ( 2 a − 7 ) = 0

( 5 b + 1 ) ( 6 b + 1 ) = 0 ( 5 b + 1 ) ( 6 b + 1 ) = 0

6 m ( 12 m − 5 ) = 0 6 m ( 12 m − 5 ) = 0

2 x ( 6 x − 3 ) = 0 2 x ( 6 x − 3 ) = 0

( 2 x − 1 ) 2 = 0 ( 2 x − 1 ) 2 = 0

( 3 y + 5 ) 2 = 0 ( 3 y + 5 ) 2 = 0

5 a 2 − 26 a = 24 5 a 2 − 26 a = 24

4 b 2 + 7 b = −3 4 b 2 + 7 b = −3

4 m 2 = 17 m − 15 4 m 2 = 17 m − 15

n 2 = 5 n − 6 n 2 = 5 n − 6

7 a 2 + 14 a = 7 a 7 a 2 + 14 a = 7 a

12 b 2 − 15 b = −9 b 12 b 2 − 15 b = −9 b

49 m 2 = 144 49 m 2 = 144

625 = x 2 625 = x 2

16 y 2 = 81 16 y 2 = 81

64 p 2 = 225 64 p 2 = 225

121 n 2 = 36 121 n 2 = 36

100 y 2 = 9 100 y 2 = 9

( x + 6 ) ( x − 3 ) = −8 ( x + 6 ) ( x − 3 ) = −8

( p − 5 ) ( p + 3 ) = −7 ( p − 5 ) ( p + 3 ) = −7

( 2 x + 1 ) ( x − 3 ) = −4 x ( 2 x + 1 ) ( x − 3 ) = −4 x

( y − 3 ) ( y + 2 ) = 4 y ( y − 3 ) ( y + 2 ) = 4 y

( 3 x − 2 ) ( x + 4 ) = 12 x ( 3 x − 2 ) ( x + 4 ) = 12 x

( 2 y − 3 ) ( 3 y − 1 ) = 8 y ( 2 y − 3 ) ( 3 y − 1 ) = 8 y

20 x 2 − 60 x = −45 20 x 2 − 60 x = −45

3 y 2 − 18 y = −27 3 y 2 − 18 y = −27

15 x 2 − 10 x = 40 15 x 2 − 10 x = 40

14 y 2 − 77 y = −35 14 y 2 − 77 y = −35

18 x 2 − 9 = −21 x 18 x 2 − 9 = −21 x

16 y 2 + 12 = −32 y 16 y 2 + 12 = −32 y

16 p 3 = 24 p 2 – 9 p 16 p 3 = 24 p 2 – 9 p

m 3 − 2 m 2 = − m m 3 − 2 m 2 = − m

2 x 3 + 72 x = 24 x 2 2 x 3 + 72 x = 24 x 2

3 y 3 + 48 y = 24 y 2 3 y 3 + 48 y = 24 y 2

36 x 3 + 24 x 2 = −4 x 36 x 3 + 24 x 2 = −4 x

2 y 3 + 2 y 2 = 12 y 2 y 3 + 2 y 2 = 12 y

For the function, f ( x ) = x 2 − 8 x + 8 , f ( x ) = x 2 − 8 x + 8 , ⓐ find when f ( x ) = −4 f ( x ) = −4 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = x 2 + 11 x + 20 , f ( x ) = x 2 + 11 x + 20 , ⓐ find when f ( x ) = −8 f ( x ) = −8 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = 8 x 2 − 18 x + 5 , f ( x ) = 8 x 2 − 18 x + 5 , ⓐ find when f ( x ) = −4 f ( x ) = −4 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = 18 x 2 + 15 x − 10 , f ( x ) = 18 x 2 + 15 x − 10 , ⓐ find when f ( x ) = 15 f ( x ) = 15 ⓑ Use this information to find two points that lie on the graph of the function.

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

f ( x ) = 9 x 2 − 4 f ( x ) = 9 x 2 − 4

f ( x ) = 25 x 2 − 49 f ( x ) = 25 x 2 − 49

f ( x ) = 6 x 2 − 7 x − 5 f ( x ) = 6 x 2 − 7 x − 5

f ( x ) = 12 x 2 − 11 x + 2 f ( x ) = 12 x 2 − 11 x + 2

Solve Applications Modeled by Quadratic Equations

The product of two consecutive odd integers is 143. Find the integers.

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The width of the carport is five feet less than twice its length. Find the width and the length of the carport.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function h ( t ) = −16 t 2 + 32 t h ( t ) = −16 t 2 + 32 t models the height, h , of the rocket above the ground as a function of time, t . Find:

ⓐ the zeros of this function, which tell us when the rocket will be on the ground. ⓑ the time the rocket will be 16 feet above the ground.

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 48 h ( t ) = −16 t 2 + 32 t + 48 models the height, h , of the ball above the ground as a function of time, t . Find:

ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 48 feet above the ground. ⓒ the height the ball will be at t = 1 t = 1 seconds which is when the ball will be at its highest point.

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/6-5-polynomial-equations

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Solving Polynomials

Solving Factoring Examples

The general technique for solving bigger-than-quadratic polynomials is pretty straightforward, but the process can be time-consuming.

Note: The terminology for this topic is often used carelessly. Technically, one "solves" an equation, such as "(polynomal) equals (zero)"; one "finds the roots" of a function, such as "( y ) equals (polynomial)". On this page, regardless of how the topic is framed, the point will be to find all of the solutions to "(polynomial) equals (zero)", even if the question is stated differently, such as "Find the roots of ( y ) equals (polynomial)".

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The first step in finding the solutions of (that is, the x -intercepts of, plus any complex -valued roots of) a given polynomial function is to apply the Rational Roots Test to the polynomial's leading coefficient and constant term, in order to get a list of values that might possibly be solutions to the related polynomial equation. Your hand-in work is probably expected to contain this list, so write this out neatly.

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You can follow this up with an application of Descartes' Rule of Signs , if you like, to narrow down which possible zeroes might be best to check. On the other hand, if you've got a graphing calculator you can use, it's easy to do a graph. The x - intercepts of the graph are the same as the (real-valued) zeroes of the equation. Seeing where the line looks as though it crosses the x -axis can quickly narrow down your list of possible zeroes that you'll want first to check.

Once you've found an x -value that you want to test, you then use synthetic division to see if you can get a zero remainder. If you do get a zero remainder, then you've not only found a zero of the original polynomial, but you've also reduced your polynomial by one degree, by effectively removing one factor.

Remember that synthetic division is, among other things, a form of polynomial division, so checking if x  =  a is a solution to "(polynomial) equals (zero)" is the same as dividing the linear factor x  −  a out of the related polynomial function "( y ) equals (polynomial)".

This also means that, after a successful division, you've also successfully taken a factor out. You should not then return to the original polynomial for your next computation for finding the other zeroes. You should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.

(This method will be demonstrated in the examples below.)

You should not be surprised to see some complicated solutions to your polynomials (that is, solutions containing square roots or complex numbers, or both); these zeroes will come from applying the Quadratic Formula to (what is usually) the final (quadratic) factor of your polynomial. You should expect that the answers will be messy.

Here's how the process plays out in practice:

Find all the zeroes of: y = 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x + 24

First, I'll apply the Rational Roots Test—

Wait. Actually, the first thing I'll do is apply a trick I've learned. First, I'll check to see if either x  = 1 or x  = −1 is a root.

(These are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because (a) it's amazing how often one of these is a zero, and (b) you can just look at the powers and the numbers to figure out if either works, because of how 1 and −1 simplify.)

When x  = 1 , the polynomial evaluates as:

2 + 3 − 30 − 57 − 2 + 24 = −60

This isn't equal to zero, so x  = 1 isn't a root. But when x  = −1 , I get:

−2 + 3 + 30 − 57 + 2 + 24 = 0

This time, it did equal zero, so now I know that x  = −1 is a root, and I can take "prove" this (in my hand-in work) by using synthetic division:

The last line of this division shows me with the new, smaller polynomial equation I'm working with now:

2 x 4 + x 3 − 31 x 2 − 26 x + 24 = 0

(I'd started with a degree-five polynomial. Since I've effectively divided out the factor x  + 1 , I've reduced the degree of the polynomial by 1 . That's how I know the last line of the division represents a degree-four polynomial.)

I've taken care of checking the two easiest zeroes. Now I'll apply the Rational Roots Test to what's left in order to get a list of potential zeroes to try:

From experience (mostly by having worked extra homework problems), I've learned that most of these exercises have their zeroes somewhere near the middle of the list, rather than at the extremes. This isn't always true, of course, but it's usually better to stay away from the larger numbers, at least when I'm getting started.

So, in this case, I won't start off by trying stuff like x  = −24 or x  = 12 . Instead, I'll start out with smaller values like x  = 2 .

And I can narrow down my options further by "cheating" and looking at the graph:

This is a fourth-degree polynomial, so it has, at most, four x -intercepts, and I can see all four of them on the graph. This means that I won't have any complex-valued zeroes.

It also looks like there may be zeroes near −1.5 and 0.5 . But the clearest solution looks to be at x  = 4 and since whole numbers are easier to work with than fractions, x  = 4 would probably be a good next value to try:

The zero remainder (at the far right of the bottom row) tells me that x  = 4 is indeed a root. And the bottom row of the synthetic division tells me that I'm now left with solving the following:

2 x 3 + 9 x 2 + 5 x − 6 = 0

Looking at the constant term " 6 " in the polynomial above, and with the Rational Roots Test in mind, I can see that the following values:

x = ±24, ±12, ±8, −4

...from my original application of the Rational Roots Test won't work for the current polynomial. Even if I didn't already know this from having checked the graph, I can see that they won't fit with the new polynomial's leading coefficient and constant term. So I can cross these values off of my list now.

(Always check the list of possible zeroes as you go. The Rational Roots Test will sometimes give a very long list of possibilities, and it can be helpful to notice that some of those values can be ignored, especially if you don't have a graphing calculator to "cheat" with.)

2 x 2 + 6 x − 4 = 0

Dividing through by 2 to get smaller numbers gives me:

x 2 + 3 x − 2 = 0

I can apply the Quadratic Formula to this:

This gives me the remaining two roots of the original polynomial function. (I plugged the exact values into my calculator, to confirm that they match up with what I'd already seen on the graph, so I'd be certain that my answer was correct. I won't hand in these approximations, though.)

My complete answer is:

Asking you to find the zeroes of a polynomial function, y equals (polynomial), means the same thing as asking you to find the solutions to a polynomial equation, (polynomial) equals (zero). The zeroes of a polynomial are the values of x that make the polynomial equal to zero. Either task may be referred to as "solving the polynomial".

So the above problem could have been stated along the lines of:

Find the solutions to 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x + 24 = 0

Find the solutions to 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x = −24

...and the answers would have been the exact same list of x -values.

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This calculator solves equations that are reducible to polynomial form , such as $ \color{blue}{2(x+1) + 3(x-1) = 5} $ , $ \color{blue}{(2x+1)^2 - (x-1)^2 = x} $ and $ \color{blue}{ \frac{2x+1}{2} + \frac{3-4x}{3} = 1} $ . The calculator will try to find an exact solution; if this is not possible, it will use the cubic or quartic formulas.

The calculator will also walk you through each step and give you a detailed explanation on how to simplify and solve the equation.

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Polynomial Equations

Polynomials are one of the significant concepts of mathematics, and so are Polynomial Equations, where the relation between numbers and variables is explained in a pattern. In math, there are a variety of equations formed with algebraic expressions. Polynomial Equations are also a form of algebraic equations.

Let us learn more about polynomial equations along with their types and the process of solving them.

What is a Polynomial Equation?

A polynomial equation is an equation where a polynomial is set equal to zero. i.e., it is an equation formed with variables , non-negative integer exponents , and coefficients together with operations and an equal sign. It has different exponents. The highest one gives the degree of the equation. For an equation to be a polynomial equation, the variable in it should have only non-negative integer exponents. i.e., the exponents of variables should be only non-negative and they should neither be negative nor be fractions. For example, 2x 2 + 3x + 1 is a polynomial and hence 2x 2 + 3x + 1 = 0 is a polynomial equation.

Here are more examples of polynomial equations:

In algebra, almost all equations are polynomial equations. Now, let's explore more details about polynomial equations.

Polynomial Equation Formula

A polynomial equation is always of the form " polynomial = 0". Algebraically, it is of the form p(x) = a n x n + a n - 1 x n - 1 + ... + a 1 x + a 0 = 0, where

• a n , a n - 1 , ...., a 1 , a 0 are coefficients and all these numbers are real numbers .
• 'x' is the variable.
• p(x) means "polynomial in terms of variable x"
• 'n' is a non-negative integer and as it is the highest exponent, it is the degree of p(x).

Polynomial Equation Examples

Here are some examples based on the polynomial equation formula.

Think: Determine a few characteristics of an algebraic equation to not to be considered as a polynomial equation.

Types of Polynomial Equations

The type of polynomial equation depends on its degree (the highest exponent of the variable). There are mainly 4 types of polynomial equations:

• Linear Polynomial Equation
• Quadratic Polynomial Equation
• Cubic Polynomial Equation
• Biquadratic Polynomial Equation

Any polynomial equation other than these is known as a higher degree polynomial equation. Let us see what each of them looks like.

Linear Equations

These are the polynomial equations with degree 1. It is of the form ax + b = 0.

Examples: 2x + 3 = 0, 5x - 7 = 0, etc.

Quadratic Equations

These are the polynomial equations with degree 2. It is of the form ax 2 + bx + c = 0.

Examples: 3x 2 - 5x + 7 = 0, x 2 + 6x + 7 = 0, etc.

Cubic Equations

These are the polynomial equations with degree 3. It is of the form ax 3 + bx 2 + cx + d = 0.

Examples: x 3 - 5 = 0, y 3 + 7y 2 - 9 = 0, etc.

Biquadratic Equations

These are the polynomial equations with degree 4. It is of the form ax 4 + bx 3 + cx 2 + dx + e = 0.

Example: 3x 4 - 5x + 2 = 0.

Solving Polynomial Equations

The process of solving polynomial equation p(x) = 0 is nothing but finding the value(s) of 'x' that satisfies the equation. A number 'a' is known as a ' zero ' of a polynomial p(x) if and only if p(a) = 0. Here, 'a' is also known as the root of the polynomial equation p(x) = 0. Hence, the process of solving polynomial equations is nothing but finding its roots.

• To know how to solve linear polynomial equations, click here .
• To know how to solve quadratic polynomial equations, click here .
• To know how to solve cubic polynomial equations, click here .

For solving any polynomials other than these, remainder theorem , factor theorem , rational root theorem , and synthetic division are very helpful. Check out each of these topics by clicking on the respective links.

Difference Between Polynomial and Equation

A polynomial is the parent term used to describe a certain type of algebraic expression that contains variables, and constants, and involves the operations of addition , subtraction , multiplication , and division along with only non-negative powers associated with the variables.

Example : 2x + 3

A polynomial equation is a mathematical statement with an ' equal to ' symbol between two algebraic expressions that have equal values.

Example : 2x + 3 = 7

Important Notes on Polynomial Equations:

• The degree of a polynomial equation is the highest power of the variable in the equation.
• Solving an equation is finding those values of the variables which satisfy the equation.
• You can also find a polynomial equation when roots are known.

☛ Related Topics:

• Polynomial Equation Solver Calculator
• Polynomial Calculator
• Polynomial Identity

Examples of Polynomial Equations

Example 1: Which of the following are polynomial equations? Justify your answers.

a) √x + 2 = 0 b) x 2 + 3x + 2 = 0 c) x/2 + 3x 2 + 5 = 0 d) 3x 3 - √2 x + 1 = 0 e) 2/(x + 3) = 0

Any equation is NOT a polynomial equation due to one of the following reasons:

• If the equation has a non-integer (or) negative exponent of the variable.
• If the equation has any variable in the denominator.

We will see whether each of the given equations is a polynomial equation or not based on these conditions.

Answer: Only b), c), and d) are polynomial equations.

Example 2: Which of the following is the polynomial equation 2x 4 - 5x 3 + 9x 2 - 4 = 0? (a) Linear Equation (b) Quadratic Equation (c) Cubic Equation (d) Biquadratic Equation.

The given polynomial equation is in terms of x. The highest power of x is 4 and hence the degree of the equation is 4. Hence, it is a biquadratic equation.

Answer: Option (d).

Example 3: Find the polynomial equation of the lowest degree in terms of x whose roots are -3 and 8.

The roots are -3 and 8. So the corresponding factors are x + 3 and x - 8. Thus, the corresponding polynomial equation is,

(x + 3) (x - 8) = 0

x 2 - 8x + 3x - 24 = 0

x 2 - 5x - 24 = 0

Answer: x 2 - 5x - 24 = 0.

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Practice Questions on Polynomial Equation

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FAQs on Polynomial Equation

How will you know if an equation is a polynomial equation.

A polynomial equation is basically a polynomial expression equated to 0. For example, 3x 2 - 5 = 0 is a polynomial equation as 3x 2 - 5 is a polynomial expression .

What is the Difference Between a Polynomial and a Polynomial Equation?

A polynomial is an expression that is made up of one or more variables, coefficients, and non-negative integer exponents of variables. An equation is a mathematical statement with an 'equal to' symbol between two algebraic expressions that have equal values. Thus, a polynomial equation is an equation that is of the form polynomial = 0.

What are the Different Types of Polynomial Equations?

The different types of polynomial equations are - linear equations , quadratic equations , cubic equations , and biquadratic equations.

What is Polynomial Equation Formula?

A polynomial formula is a polynomial function set to 0 and is of the form p(x) = a n x n + a n - 1 x n - 1 + ... + a 1 x + a 0 = 0.

What is Not a Polynomial Equation?

Any algebraic equation with a negative exponent or fractional exponent is NOT ot a polynomial equation. In other words, if an equation that has "= 0" in it doesn't have a polynomial in it, then it is NOT a polynomial equation.

What is the General Form of a Polynomial Equation?

The general form of polynomial equation in terms of x is a n x n + a n - 1 x n - 1 + ... + a 1 x + a 0 = 0. Here, a n , a n - 1 , ...., a 1 , a 0 are known as coefficients and these are real numbers.

How do You Solve Polynomial Equations?

The polynomial equations can be solved by factoring them and setting each factor to zero. Also, we can graph the left side of the polynomial equation p(x) = 0 using a graphing calculator and in that case, the x-intercepts of the graph would give the roots of the polynomial equation.

How to Find the Degree of Polynomial Equations?

The highest power of the variable term in the polynomial is the degree of the polynomial. For example, the degree of the polynomial equation x 3 + 2x + 5 = 0 is 3.

How do You Find the Roots of a Polynomial Equation?

The roots of a polynomial equation can be found using one of the following methods:

• We first find one root either by trial and error method or by using the rational root theorem. Then we use the corresponding factor and divide the given polynomial to find the other roots.
• We can find all the roots by completely factorizing the polynomial in the given equation (if possible) and by setting each factor to zero.
• We can just graph the polynomial and its x-intercepts would be the roots of the equation.

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Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. It also factors polynomials, plots polynomial solution sets and inequalities and more.

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• plot inequality x^2-7x+12<=0
• solve {3x-5y==2,x+2y==-1}
• plot inequality 3x-5y>=2 and x+2y<=-1
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About solving equations

A value c c is said to be a root of a polynomial p(x) p x if p(c)=0 p c = 0 ..

. This polynomial is considered to have two roots, both equal to 3.

One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. There are more advanced formulas for expressing roots of cubic and quartic polynomials, and also a number of numeric methods for approximating roots of arbitrary polynomials. These use methods from complex analysis as well as sophisticated numerical algorithms, and indeed, this is an area of ongoing research and development.

Systems of linear equations are often solved using Gaussian elimination or related methods. This too is typically encountered in secondary or college math curricula. More advanced methods are needed to find roots of simultaneous systems of nonlinear equations. Similar remarks hold for working with systems of inequalities: the linear case can be handled using methods covered in linear algebra courses, whereas higher-degree polynomial systems typically require more sophisticated computational tools.

How Wolfram|Alpha solves equations

For equation solving, Wolfram|Alpha calls the Wolfram Language's Solve and Reduce functions, which contain a broad range of methods for all kinds of algebra, from basic linear and quadratic equations to multivariate nonlinear systems. In some cases, linear algebra methods such as Gaussian elimination are used, with optimizations to increase speed and reliability. Other operations rely on theorems and algorithms from number theory, abstract algebra and other advanced fields to compute results. These methods are carefully designed and chosen to enable Wolfram|Alpha to solve the greatest variety of problems while also minimizing computation time.

Although such methods are useful for direct solutions, it is also important for the system to understand how a human would solve the same problem. As a result, Wolfram|Alpha also has separate algorithms to show algebraic operations step by step using classic techniques that are easy for humans to recognize and follow. This includes elimination, substitution, the quadratic formula, Cramer's rule and many more.

A polynomial is a function in one or more variables that consists of a sum of variables raised to nonnegative , integral powers and multiplied by coefficients from a predetermined set (usually the set of integers; rational , real or complex numbers; but in abstract algebra often an arbitrary field ). Note that a constant is also a polynomial.

For example, these are polynomials:

• 1.1 A More Precise Definition
• 1.2 The Degree of a Polynomial
• 1.3.1 What is a root?
• 1.3.2 The Fundamental Theorem of Algebra
• 1.3.3 Factoring
• 1.3.4 The Rational Root Theorem
• 1.3.5 Descarte's Law of Signs
• 1.4 Binomial Theorem
• 1.5 Special Values
• 2 Intermediate Topics
• 3 Olympiad Topics

Introductory Topics

A more precise definition.

The Degree of a Polynomial

The degree, together with the coefficient of the largest term, provides a surprisingly large amount of information about the polynomial: how it behaves in the limit as the variable grows very large (either in the positive or negative direction) and how many roots it has.

Finding Roots of Polynomials

What is a root.

The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that any polynomial with complex coefficients can be written as

Different methods of factoring can help find roots of polynomials. Consider this polynomial:

This polynomial easily factors to:

Now, the roots of the polynomial are clearly -3, -2, and 2.

The Rational Root Theorem

Descarte's Law of Signs

Binomial Theorem

The Binomial Theorem can be very useful for factoring and expanding polynomials.

Special Values

Intermediate Topics

• Complex numbers
• Fundamental Theorem of Algebra
• Roots of unity

Olympiad Topics

• Vieta's formulas
• Newton's identities
• Newton sums
• Polynomials

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Polynomial Equations

Polynomial equations are one of the significant concepts of Mathematics, where the relation between numbers and variables are explained in a pattern. In Maths, we have studied a variety of equations formed with algebraic expressions. When we talk about polynomials, it is also a form of the algebraic equation.

What is a Polynomial Equation?

The equations formed with variables, exponents and coefficients are called as polynomial equations. It can have different exponents, where the higher one is called the degree of the equation. We can solve polynomials by factoring them in terms of degree and variables present in the equation.

A polynomial function is an expression which consists of a single independent variable, where the variable can occur in the equation more than one time with different degree of the exponent. Students will also learn here how to solve these polynomial functions. The graph of a polynomial function can also be drawn using turning points, intercepts, end behaviour and the Intermediate Value Theorem.

Example of polynomial function:

f(x) = 3x 2  + 5x + 19

Read More:  Polynomial Functions

Polynomial Equations Formula

Usually, the polynomial equation is expressed in the form of a n (x n ). Here a is the coefficient, x is the variable and n is the exponent. As we have already discussed in the introduction part, the value of exponent should always be a positive integer.

If we expand the polynomial equation we get;

F(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + …….. + a 1 x +a 0  = 0

This is the general expression and it can also be expressed as;

Example of a polynomial equation is: 2x 2 + 3x + 1 = 0, where 2x 2 + 3x + 1 is basically a polynomial expression which has been set equal to zero, to form a polynomial equation.

Types of Polynomial Equation

A polynomial equation is basically of four types;

• Monomial Equations
• Binomial Equations
• Trinomial or Cubic Equations
• Linear Polynomial Equations
• Quadratic Polynomial Equations
• Cubic Polynomial Equation

Monomial Equation:

An equation which has only one variable term is called a Monomial equation. This is also called a linear equation . It can be expressed in the algebraic form of;

For Example:

• 8z – 3 = 0

Binomial Equations:

An equation which has only two variable terms and is followed by one variable term is called a Binomial equation. This is also in the form of the quadratic equation . It can be expressed in the algebraic form of;

ax 2 + bx + c = 0

• 2x 2 + 5x + 20 = 0
• 3x 2 – 4x + 12 = 0

Trinomial Equations:

An equation which has only three variable terms and is followed by two variable and one variable term is called a Trinomial equation. This is also called a cubic equation . In other words, a polynomial equation which has a degree of three is called a cubic polynomial equation or trinomial polynomial equation.

Since the power of the variable is the maximum up to 3, therefore, we get three values for a variable, say x.

It is expressed as;

a 0 x 3 + a 1 x 2 + a 2 x + a 3 = 0, a ≠ 0

ax 3 + bx 2 + cx + d = 0

• 3x 3 + 12x 2 – 8x – 10 = 0
• 9x 3 + 5x 2 – 4x – 2 = 0

To get the value of x, we generally use, trial and error method, in which we start putting the value of x randomly, to get the given expression as 0. If for both sides of the polynomial equation, we get 0 ,then the value of x is considered as one of its roots. After that we can find the other two values of x.

Let us take an example:

Problem: y 3 – y 2 + y – 1 = 0 is a cubic polynomial equation. Find the roots of it.

Solution: y 3 – y 2 + y – 1 = 0 is the given equation.

By trial and error method, start putting the value of x.

If y = -1, then,

(-1) 3 – (-1) 2 -1 + 1 = 0

-1 – 1 – 1 – 1 = 0

If y = 1, then,

1 3 – 1 2 + 1 – 1 = 0

Therefore, one of the roots is 1.

(y – 1) is one of the factors.

Now dividing the given equation with (y – 1), we get,

(y – 1) (y 2  +  1) = 0

Therefore, the roots are y = 1 which is a real number and y 2 + 1 gives complex numbers or imaginary numbers.

Quadratic Polynomial Equation

A polynomial equation which has a degree as two is called a quadratic equation . The expression for the quadratic equation is:

ax 2 + bx + c = 0 ; a ≠ 0

Here, a,b, and c are real numbers. The roots of quadratic equations will be two values for the variable x. These can be found by using the quadratic formula as:

Also Check: Polynomial Equation Solver

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How to Solve Polynomials

Last Updated: January 22, 2024 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 12 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 347,884 times.

A polynomial is an expression made up of adding and subtracting terms. A terms can consist of constants, coefficients, and variables. When solving polynomials, you usually trying to figure out for which x-values y=0. Lower-degree polynomials will have zero, one or two real solutions, depending on whether they are linear polynomials or quadratic polynomials. These types of polynomials can be easily solved using basic algebra and factoring methods. For help solving polynomials of a higher degree, read Solve Higher Degree Polynomials .

Solving a Linear Polynomial

Solving a Quadratic Polynomial

Community Q&A

• Remember the order of operations while you work -- First work in the parenthesis, then do the multiplication and division, and finally do the addition and subtraction. [17] X Research source Thanks Helpful 0 Not Helpful 0
• Don't fret if you get different variables, like t, or if you see an equation set to f(x) instead of 0. If the question wants roots, zeros, or factors, just treat it like any other problem. Thanks Helpful 3 Not Helpful 3

You Might Also Like

• ↑ https://www.cuemath.com/algebra/linear-polynomial/
• ↑ https://www.math.utah.edu/~wortman/1050-text-calp.pdf
• ↑ https://www.mathsisfun.com/algebra/polynomials-solving.html
• ↑ David Jia. Academic Tutor. Expert Interview. 7 January 2021.
• ↑ http://www.mathwords.com/c/constant.htm
• ↑ https://www.cuemath.com/algebra/factorization-of-quadratic-polynomials/
• ↑ http://www.themathpage.com/aprecalc/quadratic-equation.htm#double
• ↑ https://www.math.utah.edu/~wortman/1050-text-qp.pdf
• ↑ https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring
• ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-grouping/a/factoring-by-grouping
• ↑ https://content.byui.edu/file/b8b83119-9acc-4a7b-bc84-efacf9043998/1/Math-1-6-1.html

About This Article

To solve a linear polynomial, set the equation to equal zero, then isolate and solve for the variable. A linear polynomial will have only one answer. If you need to solve a quadratic polynomial, write the equation in order of the highest degree to the lowest, then set the equation to equal zero. Rewrite the expression as a 4-term expression and factor the equation by grouping. Rewrite the polynomial as 2 binomials and solve each one. If you want to learn how to simplify and solve your terms in a polynomial equation, keep reading the article! Did this summary help you? Yes No

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NEW CHAPTER 9 Polynomials

9.7: solving polynomial equations using the zero product principle, learning outcomes.

• Use the zero product  principle to solve polynomial equations
• Zero product principle : if two or more factors are multiplied to a product of zero, at least one of the factors is zero

The Zero Product Principle

If we multiply two numbers together and get an answer of zero, what can we say about the two numbers? The only way to get a product of zero is if we multiply by [latex]0[/latex]. This means that one of the factors  has  to be zero. This idea is called the zero product principle , and it is useful for solving certain kinds of equations.

Zero Product PRINCIPLE  

The Zero Product Principle states that if the product of two or more factors is [latex]0[/latex], then at least one of the factors must be [latex]0[/latex]. 

If [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or [latex]\text{both } a\text{ and }b=0[/latex].

When we say  [latex]a=0[/latex] or [latex]b=0[/latex], [latex]\text{both } a\text{ and }b=0[/latex] is implied. i.e. at least one of the factors must equal zero.

The zero product principle can be used to solve factored equations that are equal to zero.

Use the zero product principle  to solve [latex]5y=0[/latex]

By the zero product  principle, when two factors are multiplied and the result is zero at least one of them is equal to zero. Therefore, either [latex]5=0[/latex], or [latex]y=0[/latex].

In this case, we know that [latex]5[/latex] is not equal to zero, so [latex]y[/latex] must be equal to zero.

We can verify this with algebra.

[latex]\begin{array}{c}5y=0\\\text{}\\\frac{5y}{5}=\frac{0}{5}\\\text{}\\y=0\end{array}[/latex]

[latex]y=0[/latex]

We can extend this idea to products of more than just two factors.

Solve [latex]5x(x-4)(3x+2)=0[/latex]

There are three factors with a product of zero, so at least one of the factors must equal zero:

[latex]\begin{equation}\begin{aligned}5x(x-4)(3x+2)&=0 \\ 5x=0\text{ or }x-4=0\text{ or }3x+2&=0 \\ x=0\text{ or }x=4\text{ or }3x&=-2 \\x=0\text{ or }x=4\text{ or }x&=-\frac{2}{3} \end{aligned}\end{equation}[/latex]

[latex]x=0\text{ or }x=4\text{ or }x=-\frac{2}{3}[/latex]

1. Solve the equation: [latex]-7y(3y-2)(4y+1)=0[/latex]

2. Solve the equation:  [latex]x^2(2x+1)(5x-2)(x+2)=0[/latex]

Let’s consider the equation [latex]t(5-t)=0[/latex]. If we set each factor to zero and solve, we get two solutions [latex]t=0[/latex] or [latex]t=5[/latex].

Why don’t we just use the distributive property and the properties of equations to solve this kind of equation? Let’s try using the distributive property on this example to explain why this can be problematic.

[latex]\begin{equation}\begin{aligned}t\left(5-t\right)&=0 \\ 5t-t^2&=0 \\5t&=t^2 \\ \frac{5t}{t}&=\frac{t^2}{t} \\ 5&=t \end{aligned}\end{equation}[/latex]

Wait, our original solution was [latex]t=0\text{ or }t=5[/latex].  How did we lose one of the solutions?  The problem occurred when we divided both sides of the equation by [latex]t[/latex].  Remember, that we can divide both sides of an equation by the same  non-zero term. So if we want to divide by [latex]t[/latex], we must ensure that  [latex]t\ne 0[/latex]. SInce we don’t know whether[latex]t=0[/latex] or not, we cannot divide by [latex]t[/latex]. We lost the solution  [latex]t=0[/latex]  when we divided by [latex]t[/latex].

When we are solving polynomial equations, we need to use some different methods than we used to solve linear equations to make sure we get all of the correct answers.  The zero product principle is one tool that allows us to do this.

The following video presents more examples of how to use the zero product principle to solve polynomial equations that are in factored form.

Solving Polynomial Equations

In this section we will solve polynomial equations that can be factored using the zero product principle.

We will begin with an example where the polynomial is already equal to zero.

Solve:  [latex]-t^2+t=0[/latex]

To solve this equation, we need to factor the left side. Each term has a common factor of [latex]t[/latex] and the leading term is negative, so we can factor out [latex]-t[/latex] and use the zero product principle:

[latex]\begin{equation}\begin{aligned}-t^2+t&=0 \\-t(t-1)&=0\end{aligned}\end{equation}[/latex]

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.

[latex]\begin{equation}\begin{aligned}-t(t-1)&=0 \\ -t&=0\text{ OR }t-1=0 \\ t&=0\text{ OR }t=1\end{aligned}\end{equation}[/latex]

[latex]t=0\text{ OR }t=1[/latex]

The following video presents more examples of solving equations by factoring. A polynomial of degree two is often referred to as a  quadratic . i.e. a quadratic is any polynomial of the form [latex]ax^2+bx+c[/latex], where [latex]a, b, c[/latex] are real numbers.

When we don’t have a zero on one side of the equation, we can subtract those terms from both sides of the equation to force a zero on one side.

Solve: [latex]6t=3t^2-12t[/latex]

Our first goal is to try and see if we can use the zero product principle, since that is currently the only tool we know for solving polynomial equations. So, let’s move all the terms to one side, leaving zero on the other side.

[latex]\begin{equation}\begin{aligned}6t&=3t^2-12t \\0&=3t^2-12t-6t \\ 0&=3t^2-18t\end{aligned}\end{equation}[/latex]

We now have all the terms on the right side, and zero on the other side. Each term on the right side has a common factor of [latex]3t[/latex], so we can factor and use the zero product principle:

[latex]\begin{equation}\begin{aligned} 0&=3t^2-18t \\ 0&=3t(t-6)\end{aligned}\end{equation}[/latex]

Set each factor to zero and solve the equations:

[latex]\begin{equation}\begin{aligned}0&=3t(t-6) \\3t&=0\text{ OR }t-6=0 \\ t&=0\text{ OR }t=6\end{aligned}\end{equation}[/latex]

[latex]t=6\text{ OR }t=0[/latex]

The video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.

We will work through one more example that is similar to the ones above, except this example has fractions.  If we were asked to factor an expression containing fractions, we would have no choice but to work with the fractions and pull out a GCF that is a fraction. But, since an equation has two sides, we can eliminate the fractions by multiplying the whole equation by the least common multiple. Let’s do the same example both ways to see the difference.

Solve [latex]\frac{1}{2}y=-4y-\frac{1}{2}y^2[/latex]

To work with the fractions, we first find a common denominator, then factor:

[latex]\begin{equation}\begin{aligned}\frac{1}{2}y&=-4y-\frac{1}{2}y^2\\0&=-\frac{1}{2}y-4y-\frac{1}{2}y^2\end{aligned}\end{equation}[/latex]

To combine like terms, we use the common denominator [latex]2[/latex]:

[latex]\begin{equation}\begin{aligned}0&=-\frac{1}{2}y-4y-\frac{1}{2}y^2\\\text{ }\\0&=-\frac{1}{2}y-\frac{8y}{2}-\frac{1}{2}y^2\\\text{}\\0&=-\frac{9}{2}y-\frac{1}{2}y^2\end{aligned}\end{equation}[/latex]

Find the greatest common factor of the terms of the polynomial:

Factors of [latex]-\frac{9}{2}y[/latex] are [latex]-\frac{1}{2}\cdot{3}\cdot{3}\cdot{y}[/latex]

Factors of [latex]-\frac{1}{2}y^2[/latex] are [latex]-\frac{1}{2}\cdot{y}\cdot{y}[/latex]

Both terms have [latex]-\frac{1}{2}\text{ and }y[/latex] in common.

Rewrite each term as the product of the GCF and the remaining terms:

[latex]-\frac{9}{2}y=-\frac{1}{2}y\left(3\cdot{3}\right)=-\frac{1}{2}y\left(9\right)[/latex]

[latex]-\frac{1}{2}y^2=-\frac{1}{2}y\left(y\right)[/latex]

Rewrite the polynomial equation using the factored terms in place of the original terms. Remember to pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\left(9+y\right)[/latex]

[latex]\begin{equation}\begin{aligned}0&=-\frac{9}{2}y-\frac{1}{2}y^2\\\text{}\\0&=-\frac{1}{2}y\left(9\right)-\frac{1}{2}y\left(y\right)\\\text{}\\0&=-\frac{1}{2}y\left(9+y\right)\end{aligned}\end{equation}[/latex]

Solve the two equations.

[latex]\begin{equation}\begin{aligned}-\frac{1}{2}y&=0\text{ or }y+9=0\\ y&=0\text{ or }y=-9\end{aligned}\end{equation}[/latex]

[latex]y=0\text{ or }y=-9[/latex]

In this last example, we used many skills to solve one equation.  Let’s summarize them:

• We needed a common denominator to combine the like terms [latex]-4y\text{ and }-\frac{1}{2}y[/latex], after we moved all the terms to one side of the equation
• We found the GCF of the terms [latex]-\frac{9}{2}y\text{ and }-\frac{1}{2}y^2[/latex]
• We used the GCF to factor the polynomial [latex]-\frac{9}{2}y-\frac{1}{2}y^2[/latex]
• We used the zero product principle to solve the polynomial equation [latex]0=-\frac{1}{2}y\left(9+y\right)[/latex]

Sometimes solving an equation requires the combination of many algebraic principles and techniques.  The last facet of solving the polynomial equation [latex]\frac{1}{2}y=-4y-\frac{1}{2}y^2[/latex] that we should talk about is negative signs.

What follows is the same example completed by first eliminating the fractions by multiplying the whole equation by the least common multiple.

Solve the equation:  [latex]\frac{1}{2}y=-4y-\frac{1}{2}y^2[/latex]

First multiply both sides of the equation by the LCD [latex]2[/latex]:  [latex]2\cdot\frac{1}{2}y=2\cdot \left (-4y-\frac{1}{2}y^2\right )\\y=-8y-y^2[/latex]

Add [latex]y^2+8y[/latex] to both sides of the equation:  [latex]y^2+8y+y=0[/latex]

Combine like terms:  [latex]y^2+9y=0[/latex]

Factor:  [latex]y(y+9)=0[/latex]

Set each factor to zero and solve the equations:  [latex]y=0\text{ or }y+9=0\\y=0\text{ or }y=-9[/latex]

We get the same solution whether we work with the fractions or eliminate the fractions. We get to choose which way we prefer to complete such problems with fractions, but generally speaking, it is more efficient to eliminate the fractions as a first step.

The following video presents another example of solving an equation with fractional coefficients using factoring and the zero product principle.

Solve the equation: 5x^2=4+2x^2

Often, there is no GCF to pull out of an equation.

Solve the equation:  [latex]3x^2-4x-4=0[/latex]

Factor the left side of the equation:

[latex]\begin{equation}\begin{aligned}3x^2-4x-4 &=0\\ac=-12\text{ and }b=-4\text{ so split }-4x\text{ into }-6x+2x\\3x^2-6x+2x-4 &=0\\3x(x-2)+2(x-2)&=0\\(3x+2)(x-2)&=0\end{aligned}\end{equation}[/latex]

Set each factor to zero and solve:

[latex]\begin{equation}\begin{aligned}3x+2&=0\text{ or }x-2=0\\3x&=-2\text{ or }x=2\\ x &=-\frac{2}{3}\text{ or }x=2\end{aligned}\end{equation}[/latex]

[latex]x=-\frac{2}{3}\text{ or }x-2=0[/latex]

Solve the equation:  [latex]9x^4-24x^2=-30x^3[/latex]

Get all non-zero terms on the same side of the equation:

[latex]\begin{equation}\begin{aligned}9x^4-24x^2&=15x^3\\9x^4+30x^3-24x^2 &=0\end{aligned}\end{equation}[/latex]

Look for a GCF in all three terms: [latex]3x^2[/latex]

[latex]\begin{equation}\begin{aligned}3x^2\left (3x^2+10x-8 \right ) &=0\;\;\;\;\;ac=-24\text{ and }b=10\text{ so split }10x\text{ into }12x-2x\\3x^2\left ( 3x^2+12x-2x-8\right ) &=0\\3x^2\left[ 3x(x+4)-2(x+4)\right ] &=0\\ 3x^2(x+4)(3x-2)&=0\end{aligned}\end{equation}[/latex]

[latex]\begin{equation}\begin{aligned}3x^2(x+4)(3x-2)&=0\\3x^2=0\text{ or }x+4 &=0\text{ or }3x-2=0\\x=0\text{ or }x &=-4\text{ or }3x=2\\x=0\text{ or }x &=-4\text{ or }x=-\frac{2}{3}\end{aligned}\end{equation}[/latex]

[latex]x=0\text{ or }x=-4\text{ or }x=-\frac{2}{3}[/latex]

Solve the equation:  [latex]42x^4-49x^3=35x^2[/latex]

[latex]x=0\text{ or }x=-\frac{1}{2}\text{ or }x=\frac{5}{3}[/latex]

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Polynomial Equations

Learning objectives.

By the end of this section, you will be able to:

Use the Zero Product Property

• Solve quadratic equations by factoring
• Solve equations with polynomial functions
• Solve applications modeled by polynomial equations

Before you get started, take this readiness quiz.

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

We will first solve some quadratic equations by using the Zero Product Property . The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

We will now use the Zero Product Property, to solve a quadratic equation .

• Set each factor equal to zero.
• Solve the linear equations.

Solve Quadratic Equations by Factoring

• Factor the quadratic expression.
• Use the Zero Product Property.
• Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form .

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

We leave the check up to you.

In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property , one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

ⓐ the zeros of the function,

ⓑ any x -intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0.

ⓐ the zeros of the function

Solve Applications Modeled by Polynomial Equations

The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

• Read the problem. Make sure all the words and ideas are understood.
• Identify what we are looking for.
• Name what we are looking for. Choose a variable to represent that quantity.
• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Solve the equation using appropriate algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a polynomial equation.

The product of two consecutive odd integers is 323. Find the integers.

There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work.

The product of two consecutive odd integers is 255. Find the integers.

The product of two consecutive odd integers is 483 Find the integers.

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

A rectangular sign has area 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

The width is 5 feet and length is 6 feet.

A rectangular patio has area 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

The length of the patio is 12 feet and the width 15 feet.

We will use this formula to in the next example.

A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

5 feet and 12 feet

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.

24 feet and 25 feet

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

ⓐ the zeros of this function which tell us when the ball hits the ground

ⓑ when the ball will be 80 feet above the ground

ⓐ the zeros of this function which tell us when the rock will hit the ocean

ⓑ when the rock will be 160 feet above the ocean.

ⓐ 5 ⓑ 0;3 ⓒ 196

ⓐ the zeros of this function which is when the penny will hit the ocean

ⓑ when the penny will be 128 feet above the ocean.

ⓐ 4 ⓑ 0;2 ⓒ 144

Access this online resource for additional instruction and practice with quadratic equations.

• Beginning Algebra & Solving Quadratics with the Zero Property

Key Concepts

• Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

Section Exercises

Practice makes perfect.

In the following exercises, solve.

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

Solve Applications Modeled by Quadratic Equations

The product of two consecutive odd integers is 143. Find the integers.

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The height of the carport is five feet less than twice its length. Find the height and the length of the carport.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

ⓐ the zeros of this function which tells us when the rocket will hit the ground. ⓑ the time the rocket will be 16 feet above the ground.

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Answers will vary.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Greatest common factor and factor by grouping.

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

Factor by Grouping

In the following exercises, factor by grouping.

Factor Trinomials

In the following exercises, factor completely using trial and error.

In the following exercises, factor.

Factor using substitution

In the following exercises, factor using substitution.

Factor Special Products

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

In each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

The product of two consecutive numbers is 399. Find the numbers.

The area of a rectangular shaped patio 432 square feet. The length of the patio is 6 feet more than its width. Find the length and width.

A ladder leans against the wall of a building. The length of the ladder is 9 feet longer than the distance of the bottom of the ladder from the building. The distance of the top of the ladder reaches up the side of the building is 7 feet longer than the distance of the bottom of the ladder from the building. Find the lengths of all three sides of the triangle formed by the ladder leaning against the building.

The lengths are 8, 15, and 17 ft.

Chapter Practice Test

In the following exercises, solve

The product of two consecutive integers is 156. Find the integers.

The area of a rectangular place mat is 168 square inches. Its length is two inches longer than the width. Find the length and width of the placemat.

The width is 12 inches and the length is 14 inches.

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Polynomial equation definition and examples

Properties and techniques in solving polynomial equations, finding the zeros of a polynomial function, practice questions, polynomial equation – properties, techniques, and examples.

Polynomial equations are equations that contain polynomials on both sides of the equation.

Since we’re dealing with polynomials and polynomial functions, make sure to check out our article on polynomial functions .

Polynomial equations such as quadratic functions are often used in modeling motions, real-world functions, and extensive technology and science applications. This is also why we need to understand how we can identify and solve polynomial equations .

What is a polynomial equation?

Polynomial equations are equations that contain polynomial expressions on both sides of the equation. Here’s the standard form of a polynomial equation.

Note that  a­ n , a n-1 , … a o can be any complex number, and the exponents can only be whole numbers for these to be considered polynomial expressions.

 Having an equal sign followed by another polynomial expression makes the polynomial equation distinct from polynomial expressions.

As can be confirmed from the equation shown above, the polynomial equation is said to be in standard form when the terms are arranged from the term with the highest power to the one with the lowest power.

Polynomial equations contain polynomial expressions, so properties of polynomial functions will still apply. In fact, the degree and the number of terms of the polynomial expression can also help us classify polynomial equations.

Let’s go ahead and take a look at the common types of polynomial equations we may encounter based on the degree:

How to solve polynomial equations?               

A great way to visualize polynomial equations is to think of it is as the result of combining different blocks. When the goal is to find the roots, solutions, or solving for the polynomial equation, we must find a way to take each block apart.

Here are some important pointers to remember when solving polynomial equations:

• If the polynomial equation is still not in its standard form, rewrite the equation so that it is in standard form : all polynomial expressions on the left side and 0 on the right.
• Simplify the polynomial equation in standard form and predict the number of zeroes or roots that the equation might have.
• If the polynomial equation is a linear or quadratic equation, apply previous knowledge to solve these types of equations .
• If the polynomial equation has a three or higher degree, start by finding one rational factor or zero .
• Take out this factor and repeat the same process until you’re left with a linear equation or a constant .
• List down all the roots or zeros.

Let’s do a quick recap of the different techniques we can apply to achieve Step 3. As mentioned, at this point, we should know how to solve linear and quadratic equations extensively. Don’t worry. This website contains a handful of resources about these two equations in case we need a quick refresher.

Linear Equations

Linear equations are polynomial equations that have a degree of 1.

Solving for solutions for this type of equation will require us to isolate the unknown variable on one side of the equation. Master your craft in solving linear equations here .

Quadratic Equations

Quadratic equations are polynomial equations with a degree of 2.

ax 2 + bx + c = 0

There are different ways we can solve quadratic equations – it mostly depends on the form of the quadratic expression on the right-hand side.

• We can factor quadratic expressions and apply the zero-property.
• Applying special algebraic properties such as the difference of two squares , perfect square trinomial properties, and completing the square .
• Lastly, we can also use the quadratic formula to find the zeroes of quadratic equations.

Polynomial Equations (with a degree of 3 or higher)

Here’s the exciting part: what if we need to find the zeros of the solutions of a polynomial equation with degrees that are 3 or higher?

Some cubic and quartic equations can be factored by grouping and be reduced to equations with a smaller degree. There are times, however, that finding the actual factors can be challenging.

Here are important properties of polynomial equations that we’ll need to understand to easily find the real zeros and roots of a polynomial equation.

Real Zeros in a Function

The num ber of real zeroes a polynomial function can have is the same value of the degree. What does this mean? If f(x) has a degree of 5, the maximum number of real zeroes it can have is 5.

Descartes’ Rule of Signs

This rule is helpful when we need to find the zeroes of a polynomial equation without its graph. What does Descartes’s Rule of Signs do? It tells us the number and position of a polynomial equation’s zeroes.

To apply this rule, we’ll need to observe the signs between the coefficients of both f(x) and f(-x). Let’s say we have f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12.

Count the number of times the coefficients switch signs, and the table below summarizes what the result means:

Let’s apply this with f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12.

f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12 f(-x) = 2x 4 + 2x 3 – 14x 2 – 2x + 12

From the sign changes in f(x), there can be 2 or 0 positive real zeros. Similarly, from f(-x), there are can also be 2 or 0 negative real zeros.

Rational Zeros Theorem

This theorem will help us narrow down the possible rational zeros of a polynomial function .  Let p contain all the factors of a n (leading term) and q contain all the factors from a o (constant term).

The possible rational zeros of the polynomial equation can be from dividing p by q, p/q . Make sure that the list contains all possible expressions for p/q in the lowest form.

Using the same example, f(x) = 2x 4 – 2x 3 – 14x 2 + 2x + 12, we have p = 2 and q = 12 . Let’s go ahead and list down all the possible rational zeros of f(x).

Does this mean f(x) has 14 rational zeros? No, this list tells us that if f(x) has rational zeros, it will come from this list. Meaning , we have reduced the possibilities to a reasonable number from an extensive range of rational numbers .

Applying the Remainder Theorem and Synthetic Division

How do we slowly find rational zeros of f(x) once we have a list of p/q? It’s time that we apply our past knowledge on the remainder theorem , factor theorem , and synthetic division . Make sure to take a quick refresher for these topics by clicking on the links.

• This means that we can try each factor listed and check for the remainder.
• If the remainder is zero, the value of p/q is a root of f(x).
• Use the resulting polynomials and repeat the same process until we have all the f(x) zeros.

Why don’t we apply what we’ve just learned to find the zeros of 2x 4 – 2x 3 – 14x 2 + 2x + 12 = 0? From the previous section, we’ve seen that f(x)’s list of possible rational zeroes is: ±1/12, ±1/6, ±1/4, ± 1/3, ±1/2, ±2/3, ±1, and  ±2.

Let’s check if x = 1 is a root of f(x) using synthetic division.

1  |  2    -2    -14       2      12

              2       0     -14    -12

_____________________________

       2      0    -14      -12     0

Since the remainder is 0, (x – 1) is a factor of f(x) and x = 1 a solution to the equation . Let’s express f(x) as a factor of (x – 1): f(x) = (x -1)(2x 3 – 14x – 12).

Use the resulting cubic expression and find a second root for the equation. Let’s try to see if x = 2 is a root of 2x 3 – 14x – 12.

-2  |  2      0     -14      -12

               -4        8        12

__________________________

        2      -4      -6         0

Hence, (x + 2) is a factor of f(x) and x = -2 is a root of the equation . Since we have a quadratic expression, we can factor the expression and solve for the two remaining zeros of the equation.

2x 2 – 4x – 6 = 0

2(x 2 – 2x – 3) = 0

x 2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x = 3, x = -1

Let’s go ahead and take note of all the zeros we have solved for our polynomial equation: x = 1, x = -2, x = 3, and x = -1 . The equation has four zeros, as we have expected since it has a degree of 4.

Apply a similar process when finding the zeros of other polynomial equations.

Here’s a guide to help you summarize and follow through the steps we might need to do when finding the zeroes of a given polynomial equation in standard form:

The guide above helps you ask the right questions to ensure we apply the best strategy when solving polynomial equations.

Why don’t we apply this by answering the questions shown below?

Given that f(x) = -2x 3 + 4x 2 – 7x – 6, how many sign changes are there in f(x) and f(-x)? Interpret the results.

We can immediately inspect f(x) for its sign changes. We have two sign changes : -2x 3 and 4x 2 and as +4x 2 and -7x.

As for f(-x), let’s go ahead and find the expression for f(-x) first.

f(-x) = 2x 3 + 4x 2 + 7x -6

From this, we can see that f(-x) has only one sign change : between 7x and -6. Using Descartes’ Rule of Sign, we can conclude that:

• When f(x) is equated to 0, the resulting equation may have two or zero positive real zeros .
• Similarly, the resulting equation may have one negative real zero .

True or False? Given that the polynomial function, g(x), has a degree of 3, the equation g(x) = 0 will always have three real zeros.

The equation g(x) = 0 will have at most three possible real zeros. This means that it may or may not have three real zeros exactly. One function to further show that the statement is not true is when g(x) = x 3 + x.

Let’s solve the equation and observe the results for x. Since the expression is still factorable, we’ll factor x out and equate x 2 + 1 to 0.

x 3 + x = 0

x(x 2 + 1) = 0

x 2 + 1 = 0

This will only be true when x = -i or x = i. This clearly shows that it is possible for g(x) to not have three real zeros despite having a degree of 3. Hence, the statement is not true.

Find the values of x that satisfies the given equation: 4x 5 – 4x 4 + 73x 2 = -18(x -1)+ 73x 3 .

The equation is still not in its standard form, so let’s go ahead and isolate all terms on the left-hand side.

4x 5 – 4x 4 – 73x 3 + 73x 2 + 18x – 18 = 0

Using the rational zeros theorem, let’s list down the possible rational zeros for the polynomial equation.

1  |  4    -4     -73       73     18     -18

               4        0      -73       0      18

____________________________________

        4      0     -73        0      18       0

Since the remainder is 0, (x – 1) is a factor of the expression and x = 1 is a solution.  Let’s go ahead and try x = 1/2 and x = -1/2 to see if they are zeros of the equation too.

1/2  |  4      0     -73        0         18

                    2        1      -36       -18 

_________________________________

         4        2     -72     -36          0

Make sure to check with resulting polynomial. We now have (x – 1)(x – 1/2)(4x 3 + 2x 2 – 72x – 36) = 0.

-1/2  |  4         2     -72     -36          

                        -2       0       36 

            4        0     -72         0

We can see that x = -1/2 and x = 1/2 are both zeros of the polynomial equation from these two consecutive synthetic divisions.

We now have (x – 1)(x – 1/2)(x + 1/2)(4x 2 – 72) = 0. Since the remaining expression is a quadratic expression, we can equate it to 0 and solve the polynomial equation’s remaining zeros.

4x 2 – 72 = 0

4(x 2 – 18) = 0

x 2 – 18 = 0

We now have five zeros for the polynomial equation (this is already the maximum number of zeros possible for a polynomial equation with a degree of 5).

Hence, the equation has a solution set of: {-2√2, -1/2, 1/2, 1, 2√2}.

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PRACTICE PROBLEMS ON SOLVING POLYNOMIAL EQUATIONS

(1)  Solve the cubic equation : 2x 3 − x 2 −18x + 9 = 0, if sum of two of its roots vanishes         Solution

(2)   Solve the equation 9x 3 − 36x 2 + 44x −16 = 0 if the roots form an arithmetic progression.         Solution

(3)  Solve the equation 3x 3  − 26x 2  + 52x − 24 = 0 if its roots form a geometric progression.         Solution

(4)  Determine k and solve the equation 2x 3  − 6x 2  + 3x + k = 0 if one of its roots is twice the sum of the other two  roots.         Solution

(5)  Find all zeros of the polynomial x 6  − 3x 5  − 5x 4  + 22x 3  − 39x 2  − 39x + 135, if it is known that 1 + 2i an d  √ 3 are two of its zeros.              Solution

(6)   Solve the cubic equation

(i) 2x 3  − 9x 2  +10x = 3

(ii)  8x 3  − 2x 2  − 7x + 3 = 0.        Solution

(7)  Solve the equation x 4  −14x 2  + 45 = 0    Solution

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Factorization of Polynomials Practice Problems

Factoring polynomials is a fundamental skill in algebra that involves breaking down a polynomial into simpler factors that, when multiplied together, give the original polynomial. This process is crucial for solving polynomial equations, simplifying expressions, and understanding polynomial functions.

The practice problems for factoring polynomials cover various techniques and methods. These include factoring out the greatest common factor (GCF), using the difference of squares, applying the quadratic formula, and grouping terms. Understanding these methods allows students to tackle more complex polynomial equations and is essential for higher-level mathematics.

What are Polynomials?

Polynomials are algebraic expressions that consist of variables and coefficients, structured as a sum of terms where each term includes a variable raised to a non-negative integer exponent.

A polynomial in one variable x is generally written as:

P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0

• a n , a n-1 , . . ., a 1 , a 0 are coefficients.
• n is a non-negative integer representing the degree of the polynomial (the highest power of the variable).
• a n is the leading coefficient, and a 0 is the constant term.

Factors of Polynomial

Factors of a polynomial are expressions that can be multiplied together to yield the original polynomial.

There are various methods for finding factors of polynomial such as:

Factoring by Grouping

This method is used when a polynomial has four or more terms. Grouping terms with common factors and then factoring out the GCF from each group simplifies the polynomial.

Example: Find factors of polynomial x 3 + 3x 2 + x + 3.

Group terms: (x 3 + 3x 2 )+ (x + 3) Factor out the GCF from each group: x 2 (x + 3) + (x + 3) Factor out the common binomial factor: (x + 3)(x 2 + 1)

Difference of Squares

A polynomial in the form a 2 − b 2 can be factored as (a + b)(a − b).

• For example: x 2 − 16 = (x + 4)(x − 4)

Sum or Difference of Cubes

A polynomial in the form a 3 + b 3 can be factored as (a + b)(a 2 − ab + b 2 ) and a 3 − b 3 as (a − b)(a 2 + ab + b 2 ).

• For example: x 3 + 8 = (x + 2)(x 2 − 2x + 4)

Some Other Identities

Theorems Related to Factorization of Polynomial

Factor Theorem

Factor Theorem states that a polynomial P(x) has a factor (x – c) if and only if P(c) = 0. In other words, c is a root (or zero) of the polynomial P(x) if and only if (x – c) is a factor of P(x).

Example: For the polynomial P(x) = x 3 – 6x 2 + 11x – 6, if P(2) = 0, then (x – 2) is a factor of P(x).

Remainder Theorem

Remainder Theorem states that the remainder of the division of a polynomial P(x) by (x – c) is P(c). This theorem is closely related to the Factor Theorem and is useful for quickly determining whether a linear binomial is a factor of a polynomial.

Example: If P(x) = x 3 – 6x 2 + 11x – 6, then the remainder when P(x) is divided by (x – 2) is P(2) = 0.

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every non-zero polynomial of degree n with complex coefficients has exactly n roots in the complex number system. This theorem guarantees the existence of at least one root for polynomials of degree n ≥ 1.

Example: The polynomial P(x) = x 4 + 2x 3 – 7x 2 + 8x – 3 has exactly four roots in the complex number system.

Polynomial Division Algorithm

Polynomial Division Algorithm states that given two polynomials P(x) and D(x), where D(x) is non-zero, there exist unique polynomials Q(x) (the quotient) and R(x) (the remainder) such that:

P(x) = D(x) · Q(x) + R(x)

Where the degree of R(x) is less than the degree of D(x).

Rational Root Theorem

Rational Root Theorem provides a method to identify possible rational roots of a polynomial equation P(x) = 0. It states that any rational root, p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

Example: For P(x) = 2x 3 – 3x 2 – 8x + 3, the possible rational roots are ± 1, ± 3, ± 1/2, ± 3/2.

Sample Problems on Factorization of Polynomial

Example 1: Factorize the polynomial: x 2 – 5x + 6.

To factorize x 2 – 5x + 6, we look for two numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of the linear term). These numbers are -2 and -3. x 2 – 5x + 6 = x 2 – 2x – 3x + 6 = x(x – 2) – 3(x – 2) = (x – 2)(x – 3)

Example 2: Factorize the polynomial: 2x 2 – 8x.

First, factor out the greatest common factor (GCF), which is 2x. 2x 2 – 8x = 2x(x – 4)

Example 3: Factorize the polynomial: x 3 – 27

Recognize that x 3 – 27 is a difference of cubes. Use the formula a 3 – b 3 = (a – b)(a 2 + ab + b 2 ), where a = x and b = 3. x 3 – 27 = (x – 3)(x 2 + 3x + 9)

Example 4: Factorize the polynomial: x 2 + 6x + 9

Recognize that x 2 + 6x + 9 is a perfect square trinomial. Use the formula (a + b) 2 = a 2 + 2ab + b 2 , where a = x and b = 3. x 2 + 6x + 9 = (x + 3) 2

Example 5: Factorize the polynomial: x 2 – 4x – 12

To factorize x 2 – 4x – 12, we look for two numbers that multiply to -12 (the constant term) and add up to -4 (the coefficient of the linear term). These numbers are -6 and 2. x 2 – 4x – 12 = (x – 6)(x + 2)

Example 6: Factorize the polynomial: x 3 + 3x 2 – 4x – 12

To factorize x 3 + 3x 2 – 4x – 12, we use the factor by grouping method. Group the terms: x 3 + 3x 2 – 4x – 12 = (x 3 + 3x 2 ) + (-4x – 12) Factor out the common factors from each group: x 2 (x + 3) – 4(x + 3) Factor out the common binomial factor (x + 3): (x + 3)(x 2 – 4) Further factorize x 2 – 4 as a difference of squares: x 2 – 4 = (x – 2)(x + 2) Thus, x 3 + 3x 2 – 4x – 12 = (x + 3)(x – 2)(x + 2)

Example 7: Factorize the polynomial: 4x 2 – 25

Recognize that 4x 2 – 25 is a difference of squares. Use the formula a 2 – b 2 = (a – b)(a + b), where a = 2x and b = 5. 4x 2 – 25 = (2x – 5)(2x + 5)

Problem 8: Find the zeros of the polynomial p(x) = x 3 – 4x 2 + x + 6.

p(x) = x 3 – 4x 2 + x + 6 Possible rational zeros are ± 1, ± 2, ± 3, ± 6. For x = 1: p(1) = 1 3 – 4(1) 2 + 1 + 6 = 1 – 4 + 1 + 6 = 4 (not a zero) For x = 2: p(2) = 2 3 – 4(2) 2 + 2 + 6 = 8 – 16 + 2 + 6 = 0 (not a zero) Using synthetic division or polynomial division: x 3 – 4x 2 + x + 6 = (x – 2)(x 2 – 2x – 3) As x 2 – 2x – 3 = (x – 3)(x + 1) Thus, x 3 – 4x 2 + x + 6 = (x – 2)(x – 3)(x + 1)

Problem 9: Find the factors of the polynomial p(x) = x 4 – 5x 3 + 6x 2 .

p(x) = x 4 – 5x 3 + 6x 2 ⇒ p(x) = x 2 (x 2 – 5x + 6) As x 2 – 5x + 6 = (x – 2)(x – 3) Thus, p(x) = x 2 (x – 2)(x – 3)

Example 10: Find all he factors of P(x) = x 3 – 6x 3 – 6x + 36.

P(x) = x 3 – 6x 3 – 6x + 36 Notice that we can group and factor by grouping: P(x) = (x 3 – 6x 3 ) – (6x – 36) Factor out the common terms: = x 2 (x – 6) – 6(x – 6) = (x – 6)(x 2 – 6)

Problem 11: Find all factors of the polynomial P(x) = 2x 3 – 3x 2 – 8x + 12.

For P(x) = 2x 3 – 3x 2 – 8x + 12 Factors of 12 (constant term): ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 Factors of 2 (leading coefficient): ± 1, ± 2 Possible rational zeros: ± 1, ± 1/2, ± 2, ± 3, ± 3/2, ± 4, ± 6, ± 12 Evaluate P(x) for each possible rational zero. For x = 1: P(1) = 2(1) 3 – 3(1) 2 – 8(1) + 12 = 2 – 3 – 8 + 12 = 3 ≠ 0 For x = -1: P(-1) = 2(-1) 3 – 3(-1) 2 – 8(-1) + 12 = -2 – 3 + 8 + 12 = 15 ≠ 0 For x = 2: P(2) = 2(2) 3 – 3(2) 2 – 8(2) + 12 = 16 – 12 – 16 + 12 = 0 Thus, x = 2 is a zero of the polynomial P(x). After identifying x = 2 as a zero, you can factor the polynomial using synthetic division or long division to find other zeros if necessary. \begin{array}{r|rrrr} 2 & 2 & -3 & -8 & 12 \\ & & 4 & 2 & -12 \\ \hline & 2 & 1 & -6 & 0 \end{array} This gives us 2x 3 – 3x 2 – 8x + 12 = (x – 2)(2x 2 + x – 6). Factor 2x 2 + x – 6 further: 2x 2 + x – 6 = (2x – 3)(x + 2) So, the polynomial can be factored as: P(x) = (x – 2)(2x – 3)(x + 2)

Practice Problems on Factorization of Polynomial

FAQs on Factorization of Polynomial

What is factorization of polynomials.

Factorization of polynomials is the process of expressing a polynomial as a product of its factors, which are polynomials of lower degrees.

What are the common methods used for factorizing polynomials?

Common methods include: Factoring out the greatest common factor (GCF) Factoring by grouping Using special algebraic identities Factoring quadratic polynomials Factoring cubic polynomials Using synthetic division or the rational root theorem

What is the greatest common factor (GCF)?

The GCF of a polynomial is the highest degree monomial that divides each term of the polynomial without leaving a remainder.

What is the factor theorem?

The factor theorem states that a polynomial f(x) has a factor (x − c) if and only if f(c)=0. This helps in finding the roots of the polynomial, which are used to factorize it.

How do you factor polynomials by grouping?

Factoring by grouping involves rearranging the terms of the polynomial and then factoring out the common factors in pairs or groups of terms.

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how to solve all forms of polynomials

Before attempting to solve a polynomial equation , you must first write the problem down in the standard form.

After it has been factored and is equal to zero, you should next set the value of each variable factor to zero .

The answers to the resultant equations are the same as the answers to the first set of equations. There are certain polynomial equations that cannot be factored.

Generally, An equation that may be expressed in the form is called a polynomial equation, and it refers to a specific kind of problem.

ax^n + bx^ (n-1) + cx^(n-2) (n-2) + ... + dx + e = 0

x is the independent variable , and

a, b, c, d, and e are the variables that remain the same.

The term "degree" refers to the value of n in a polynomial equation since n represents the largest power of x in the equation.

For instance, the form of a quadratic equation that has a degree of 2 and can be represented as

ax^2 + bx + c = 0 is as follows:

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Related Questions

what is this for brainliested

The solution of the expression 1/2 divided by 1/3 is 1.5.

The mathematical phrase combines numerical variables and operations symbolised by the signs for addition, subtraction, multiplication, and division.

The representation of variables, operations, functions, operations, brackets, punctuation, and grouping can all be done using mathematical symbols. They can also signify other characteristics of the logical grammar, such as the operation order.

Given that,

use the tape diagram to represent and find the value of 1/2 divided by 1/3.

Thus, the value of 1/2 divided by 1/3 is calculated as:

or, E = ( 1 / 2 ) / ( 1 / 3 )

or, E = 1.5

Therefore, the solution of the expression 1/2 divided by 1/3 is 1.5.

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Using the vertical line test, determine if the graph above shows a relation, a function, both a relation and a function, or neither a relation nor a function. A. function only B. neither a relation nor a function C.both a relation and a function D. relation only

By doing the vertical line test we can see that it is not a function , then the correct option is D.

Remember that a function is a relation where all the inputs are mapped into only one output.

So, if we draw any vertical line , and it intercepts a graph at more than one point, then the graph does not represent a function

Here we have a graph, and if you draw the lines:

All of these lines intercept the graph at two points, then it is not a fnction.

The correct option is C.

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100 POINTS, NEED HELP ASAP

[tex]\textsf{a)} \quad BC^2=CD \cdot CF[/tex]

[tex]\text{b)} \quad BC=12\; \sf inches[/tex]

Step-by-step explanation:

The relationship among the lengths of the segments formed by the secant, CD, and the tangent segment, BC, is that the square of the measure of the tangent segment , BC, is equal to the product of the measures of the secant segment , CD, and its external secant segment , CF.  

Therefore, the equation is:

[tex]BC^2=CD \cdot CF[/tex]

Yes, it is possible to find the length of BC .  

Using the equation above, we can substitute in the given values of CD and CF and solve for BC :

[tex]\begin{aligned}BC^2&=CD \cdot CF\\&=16 \cdot 9\\&=144 \end{aligned}[/tex]

Square root both sides of the equation:

[tex]\begin{aligned} \sqrt{BC^2}&=\sqrt{144}\\BC&=12\end{aligned}[/tex]

Therefore, the length of BC is 12 inches.

QUESTIONS IN PICTURE/ATTACHMENT:

The domain of the question is expressed as; 0 ≤ x ≤ 4

The range of the question is expressed as; 100 ≤ f(x) ≤ 207.36

The domain of a graph is defined as the set of all possible input values that makes the function possible while the range is defined as the set of all possible output values that can result from the possible input values.

Now, we are told that the insect population increases by 20% each month from May 1 to September 1.

The function that represents the insect population after x months is;

f(x) = 100(1.2)ˣ

Thus, the domain is from x = 0 to 4 months inclusive. 0 ≤ x ≤ 4

f(0) = 100(1.2)⁰

f(4) = 100(1.2)⁴

f(4) = 207.36

100 ≤ f(x) ≤ 207.36

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The display summarizes grades on a World History exam. A vertical box plot is titled world history exams, with a minimum of 54, a maximum of 94, a lower quartile of 70, a median of 83, and an upper quartile of 88. Which of the following describes the data set? The data is univariate and categorical. The data is univariate and numerical. The data is bivariate and categorical. The data is bivariate and numerical. B is wrong

The statement which correctly describes the data set include the following: D. the data is bivariate and numerical.

In Mathematics, a bivariate data can be defined as a type of data set which comprises information that are based on two variables, usually two types of related data.

In Mathematics and statistics, a numerical data can be defined as a type of data set that is primarily expressed in numbers only. This ultimately implies that, a numerical data simply refers to a type of data set consisting of numbers (numerals), rather than words or letters.

In conclusion, we can logically deduce that the given data set is both bivariate and numerical because it comprises the grades and the subject (World History exam).

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8 1/3 in 10 minutes

5/6 pages of a book in 1 minute

Help!!!!! What did I do wrong? What is the right answer?

the intercept on Y is 3

also the slope goes this direction means the slope must be negative

△QRS≅△PSR. Complete the proof that △PQS≅△QPR.

By Angle-Angle-Angle congruency , triangle △PQS≅△QPR is congruent.

Therefore, an isosceles triangle has two equal sides as well as two equal angles. The Greek words iso (same) and skelos are the source of the name (leg). An equilateral triangle is one in which all of its sides are equal, and a scalene triangle is one in which none of its sides are equal.

Given that:

QP=8cm,PR=6cm and SR=3cm

(I)  In △PQR and △SPR

∠PRQ=∠SRP  (Common angle)

∠QPR=∠PSR   (Given that)

∠PQR=∠PSR   (Properties of triangle )

∴△PQR∼△SPR  (By AAA)

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Find the equations of any vertical asymptotes. x² +7 (x²-16) (x²-1) f(x) = Find the vertical asymptote(s). Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. OA. The function has one vertical asymptote, (Type an equation.) OB. The function has two vertical asymptotes. The leftmost asymptote is and the rightmost asymptote is (Type equations.) OC. The function has three vertical asymptotes. The asymptotes in order from leftmost to rightmost are (Type equations.) O D. The function has four vertical asymptotes. The asymptotes in order from leftmost to rightmost are (Type equations.) O E. The function has no vertical asymptotes. and .. and

The correct option regarding the vertical asymptotes of the function are given as follows:

D. The function has four vertical asymptotes. The asymptotes in order from leftmost to rightmost are x = -4, x = -1, x = 1, x = 4.

The function for this problem is defined as follows:

[tex]F(x) = \frac{x^2 + 7}{(x^2 - 16)(x^2 - 1)}[/tex]

The vertical asymptotes of a function are the values of x that are outside the function domain.

In this problem, we have a fraction , hence the vertical asymptotes are the zeroes of the denominator, as follows:

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Question content area top Part 1 There are ten bird baths in a park. On the first day of​ spring, the bird baths are filled. Several weeks​ later, the overall change in the water level is found. The results are shown in the table. What is the range of the​ data?

The range of the​ given data would be 3.2 which is determined by the difference between the maximum value and the minimum value of data.

The difference between the largest value of the data and the minimum value of the data is measured as the range of the data.

The information provided in the table is:

{ 1.5, -1.1, -1.5, 2.9, 1.6, -2.5, -0.3, 0.9, 2.7, -1.8 }

Find the highest value and the smallest value of the data to determine the range of the data.

based on the table provided,

The highest value of the data is 2.9.

The smallest value of the data is -0.3.

Thus, the data's range would be as follows:

Both the data's maximum and minimum values are presented.

⇒ 2.9 - (-0.3) (-0.3)

⇒ 2.9 + 0.3

Apply the addition operation in mathematics.

Consequently, the provided data's range would be 3.2

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50%of what number is 140

Answer: 140 is 50% of 280

Step-by-step explanation: your welcome :)

You are standing 75 meters from the base of the Jin Mao Building in Shanghai, China. You estimate that the angle of elevation to the top of the building is 80°. Suppose one of your friends is at the top of the building. What is the distance of the line of sight between you and your friend?

Which describes the slope of this line? 1.) Negative Slope 2.) Zero Slope 3.) Positive Slope 4.) Undefined Slope​

Answer: 2 - Zero slope

It's a horizontal line

Find the value of X Given: Ray BD bisects angle ABC

By the definition of an angle bisector, [tex]m\angle ABD=m\angle DBC[/tex].

[tex]4x=2x+30 \implies 2x=30 \implies x=\boxed{15}[/tex]

Find the value of a. Round to the nearest tenth. 18 A a = [ ? 28° Law of Sines: a sin A = b sin B 83° = C a sin C B Enter

using the Law of Sines

[tex]\frac{a}{sinA}[/tex] = [tex]\frac{b}{sinB}[/tex] = [tex]\frac{c}{sinC}[/tex]

where a, b , c are the sides opposite the vertices A, B, C

here a is opposite ∠ A = 28°  and c = 18 is opposite ∠ C = 83° , then

[tex]\frac{a}{sinA}[/tex] = [tex]\frac{c}{sinC}[/tex]

[tex]\frac{a}{sin28}[/tex] = [tex]\frac{18}{sin83}[/tex] ( cross- multiply )

a × sin83° = 18 × sin28° ( divide both sides by sin83° )

a = [tex]\frac{18sin28}{sin83}[/tex] ≈ 8.5 ( to the nearest tenth )

The value of a is obtained as follows:

The law of sines states that the sine of each angle is proportional to the side length opposite the angle.

Applying the law of sines, the equation for this problem is given as follows:

a/sin(28º) = 18/sin(83º).

Applying cross multiplication, we have that:

a = 18 x sin(28º)/sin(83º)

Which is the desired side length for this problem.

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To be eligible for the senior admission rate, a visitor must be older than 65 years. Use y to represent the age (in years) of a visitor who gets the senior rate

In order to be eligible for the senior admission rate, a visitor must be older than 65 years - it is represented by y as y > 65.

While comparing two values when one value is greater than the other, the greater than sign can be used. It is indicated by the letter ">."

The greater than sign indicates that one number is greater than the other. When a number is marked with a larger than or equal sign, it signifies the leftmost number is more than or on par with the rightmost number. The number on the left is smaller than the number on the right if the less than sign is present. The signs for larger than and less than are additionally referred to as inequality signs. Equal does not imply unequal. While comparing two values that might not be equal, these signs are excellent.

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Which equation has the same solution as x^2-8x+19 = -10

So, (x - 4)² = -45 this equation , it has the same solution as x² - 8x + 19 = - 10.

An equation is a mathematical statement that proves two mathematical expressions are equal in algebra, and this is how it is most commonly used. In the equation 3x + 6 = 18, for instance, the two expressions 3x + 6 and 18 are separated by the 'equal' sign.

Standard form, slope-intercept form, and point-slope form are the three main types of linear equations.

The equation becomes:

x² - 8x + 19 = - 10

or, x² - 8x + 29 = 0

Now, Δ = b² - 4ac

or, Δ = (8)² - 4 × 1 × 29

or, Δ = - 52 < 0

Thus, it has no solution.

Correct option: (x - 4)² = -45

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Question 3 A total of $500 was invested, part of it at 7% interest and the ren amounted to $38, how much was invested at the 8% rate?

Amount $300 was invested at the 8% rate .

What is interest?

Interest is most frequently reflected as an periodic chance of the quantum of a loan . This chance is known as the interest rate on the loan. For illustration, a bank will pay you interest when you deposit your plutocrat in a high- yield savings regard.

The three types of interest include simple (regular) interest, accrued interest, and compounding interest.

To calculate simple interest, multiply the top quantum by the interest rate and the time. The formula written out is" Simple Interest = star x Interest Rate x Time." This equation is the simplest way of calculating interest.

Given that;

Total investement = $500

at the interest of 7%.

Let $x be the  invested at 7% and (500-x ) was invested at 8%

∴ The total interest after 1 year = $ [x *7*1/100 + (500-x) *8*1/100]

According to problem,

$ [x *7*1/100 + (500-x) *8*1/100] = $38

⇒7x +4000 - 8x = 38×100

⇒ -x = 3800 - 4000

⇒ x = $ 200

Therefore $(500-200)= $300 was invested at 8% rate.

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Solve each system of equations using graphing. Identify wether each system has one solution, no solution or infinite solutions.

The graph of the system of equations:

y = -3x + 4

Can be seen on the image at the end, there we can see that the lines are parallel, and thus, the system has no solutions.

Here we want to solve a system of equations graphically, let's do the first one.

To solve it graphically, you need to graph both of these lines on the same coordinate axis.

Remember that to graph a line you need to find two points on the line, to get that, you need to evaluate the line.

For example with the first one:

y = 5 - 3*0 = 5

y = 5 - 3*1 = 5 - 3  =2

So we have the points (0, 5)and (1, 2). Now you can graph these points and connect them with a line, that will be the graph of the line.

Once you have both graphs on the coordinate axis, you need to find the point where the graphs intercept. That point will be the solution.

The graph of the system can be seen on the image below, there, we can see that the two lines are parallel lines, thus they never do intercept, then this system of equations has no solutions.

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Using the Law of Sines to solve the all possible triangles if ZA = 109°, a = 35, b = 11. If no answer exists, enter DNE for all answers. LB is degrees; degrees; LC is C=

Using sine law , the value of B, C and c are 17.3°, 53.7° and 29.8 respectively

The sine law is a mathematical relationship between the lengths of the sides of a triangle and the sines of the angles opposite those sides. The sine law states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides and angles in the triangle. It is often used to solve triangles, particularly in cases where the length of one side and the measure of the angle opposite it are known.

The formula is given as;

a / sin A = b / sin B = c / sin C

In the question given;

35 / sin 109 = 11 / sin B

sin B = (11 * sin 109) / 35

sin B = 0.29716

B = sin⁻¹(0.29716)

Using sum of angles in a triangle

A + B + C = 180

109 + 17.3 + C = 180

C = 180 - 126.3

Using sin law

35 / sin 109 = c / sin 53.7

c = (35 * sin 53.7) / sin 109

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To find the height of a fir tree, Bill first found the length of the tree's shadow to be 80 feet long. At the same time, he held a yardstick perpendicular to the ground. The yardstick cast a shadow of 4 feet. What was the height of the tree? (Hint: How many feet are in a yardstick)

Answer: 60 ft

We can set up ratios to solve this problem. First, let's use x to represent the actual height of the tree. The tree, with heigh x, casts a shadow of 80 feet.

This gives us a ratio of 80:x

Next, they say that a yardstick (1 yd = 3 ft) casts a shadow of 4 ft. Therefore we can set up a ratio of 4:3.

We can compare these ratios by saying

80/x = 4/3. By cross multiplying, you get that 4x = 240.

Show that 0-52200-48826-5 is an invalid UPC number.

The total of the even and odd positions digits yields 82, which does not terminate with 0 and is therefore not a UPC number.

A UPC is examined in the manner described below: Add up the values in the first, third, fifth, seventh, ninth, and eleventh positions. The average of the digits mostly in places of 2, 4, 6, 8, 10, and 12 should be added to this. The code will be recognised by the computer as valid if the outcome is a multi of 10. Every version of your offering will often require its own UPC. Each distinct colour or pattern of your product, if it has either, will require a UPC. Each size of your product will require a UPC if it comes in various sizes. The same is true for various packing and setups.

0-52200-48826-5

= 3( 0+ 2+0+4+8+6)+ ( 5+2+0+8+2+5)

= 3(20)+ 22 = 60+22

hence it is not UPC number as it ends by 2

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What does 10 km equal to

Pleaseee help I have a stat test pleaseee

The correct statements are given as follows:

7. E. A parameter refers to the population and a statistic refers to the sample.

8. A. [tex]\sigma = 16[/tex]

9. C. Approximately normal with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{4}[/tex]

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For item 8 , the population standard deviation is then obtained as follows:

[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

[tex]2 = \frac{\sigma}{\sqrt{64}}[/tex]

[tex]2 = \frac{\sigma}{8}[/tex]

[tex]\sigma = 16[/tex]

For item 9 , the sampling distribution is approximately normal, while the sample mean is the same as the population mean, with standard deviation given as follows:

[tex]s = \frac{\sigma}{\sqrt{16}} = \frac{\sigma}{4}[/tex]

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The length of rectangle is 5 more than the width. The area is 150 square inches. What is the length and width of the rectangle?

15 length 10 width.

15 x 10 is 150.

rolls are being prepared to go to grocery stores. Divide 48 rolls into 2 groups so the ratio is 3 to 5. ​

We get 18 rolls and 30 rolls when we divide 48 rolls into two groups with a ratio of 3 to 5.

Describe ratio.

The ratio concept is used in mathematics when comparing two or more numbers.

It works as a comparison tool to demonstrate how big or small a particular quantity is.

A ratio divides two amounts in order to compare them. The "antecedent" in this case is the dividend, and the "consequent" is the divisor.

Given that there are 48 rolls altogether.

Let the first group of rolls be 3x.

Let the second group of rolls be 5x.

So, 3x + 5x = 48

Now, we find

Thus, the first group has 18 rolls.

Next, 5x = 5(6)

Thus, the second group has 30 rolls.

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Rolls are being prepared to go to grocery stores. Divide 48 rolls into 2 groups so the ratio is 3 to 5.

Determine the number of balls

Anna would like to purchase a new bike that cost $205. She already has saved $34. If she saves a maximum of $12 a week, the following inequality can be used to find the minimum number of weeks, w, it will take Anna to save the money to purchase the bike Please show steps

nut gut 20 butt 12 revolver honey bun

It will take at least 15 weeks

12w+31 ≥ 205

Subtract 31 from each side

12w+31-31 ≥ 205-31

Divide each side by 12

12w/12 ≥ 174/12

Rounding up to the nearest integer

HELP!!!!!! WHAT DID I do wrong????

Y = 1/4x + 0 or Y = 1/4x

Teacher circles the 4 because the point was (1,4)

Slope is rise over run

You risen 1 unit on the y-axis, then you went over 4 units on the axis, making the slope 1/4x

I need help with this problem

Answer: B equation C only

36.00-25.95=t

The path of two bumper cars can be represented by the functions x + y = –2 and y = x2 – x – 6. At which locations will the bumper cars hit one another? (–1, –4) and (1, –6) (–2, 0) and (2, –4) (–2, 0) and (1, –6) (–1, –4) and (2, –4)

For given equation of line and parabola for two bumper cars they will hit one another at (–2, 0) and (2, –4) i.e. B .

What is parabola ?

A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point (known as the focus ) and from a fixed straight line which is known as the directrix .

In general, if the directrix is parallel to the y-axis in the standard equation of a parabola is given as:

If the parabola is sideways i.e., the directrix is parallel to x-axis, the standard equation of a parabola becomes,

Apart from these two, the equation of a parabola can also be y2 = -4ax and x2 = -4ay if the parabola is in the negative quadrants .

After making graph for given equations as we can see in image

that they intersect each other at (–2, 0) and (2, –4).

           Two bumper cars they will hit one another at (–2, 0) and (2, –4).

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