problem solving polynomial equation

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:

Example: f(x) = (x−2) 2 (x−4) 3

(x−2) has even multiplicity , so it just touches the axis at x=2

(x−4) has odd multiplicity , so it crosses the axis at x=4

• We can directly solve polynomials of Degree 1 (linear) and 2 (quadratic)
• For Degree 3 and up, graphs can be helpful
• Know how far left or right the roots may be
• Know how many roots (the same as its degree)
• Estimate how many may be complex, positive or negative

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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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 339,163 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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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.

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• 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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About solving equations

A value is said to be a root of a polynomial if ..

The largest exponent of appearing in is called the degree of . If has degree , then it is well known that there are roots, once one takes into account multiplicity. To understand what is meant by multiplicity, take, for example, . 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.

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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• Book title: Intermediate Algebra 2e
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Polynomial Equation Word Problems (Mixed Operations)

These lessons help Algebra students learn how to write and solve polynomial equations for algebra word problems.

Related Pages Solving Challenging Word Problems Math Word Problems More Algebra Lessons

How To Solve Polynomial Equation Word Problem?

Example: A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

Polynomial Equation Word Problem

Example: A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air as follows: h = -16t 2 + 8t + 8 a) How long will it take the gymnast to reach the ground? b) When will the gymnast be 8 feet above the ground?

How To Solve Word Problems With Polynomial Equations?

• The sum of a number and its square is 72. Find the number.
• The area of a triangle is 44m 2 . Find the lengths of the legs if one of the legs is 3m longer than the other leg.
• The top of a 15-foot ladder is 3 feet farther up a wall than the foo is from the bottom of the wall. How far is the ladder from the bottom of the wall?
• A projectile is launched upward from ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground?

How To Write Polynomials For Word Problems?

Learn to write a polynomial for Word problems involving perimeter and area of rectangles and circles.

Learn How To Write And Solve Polynomial Equations

Learn to write and solve polynomial equations for special integers, consecutive integers.

Example 1: Find a number that is 56 less than its square. Let n be the number. Example 2: Find two consecutive odd integers whose sum is 130.

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4.4: Solve Polynomial Equations by Factoring

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Learning Objectives

• Review general strategies for factoring.
• Solve polynomial equations by factoring.
• Find roots of a polynomial function.
• Find polynomial equations given the solutions.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials.

general guidelines for factoring polynomials

Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Step 2: Determine the number of terms in the polynomial.

• Factor four-term polynomials by grouping.
• Factor trinomials (3 terms) using “trial and error” or the AC method.
• Difference of squares :\(a^{2}−b^{2}=(a+b)(a−b)\)
• Sum of squares : \(a^{2}+b^{2}\)  no general formula
• Difference of cubes : \(a^{3}−b^{3}=(a−b)(a^{2}+ab+b^{2})\)
• Sum of cubes : \(a^{3}+b^{3}=(a+b)(a^{2}−ab+b^{2})\)

Step 3: Look for factors that can be factored further.

Step 4: Check by multiplying.

If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Example \(\PageIndex{1}\):

Factor \(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x\).

This four-term polynomial has a GCF of \(2x\). Factor this out first.

\(54 x ^ { 4 } - 36 x ^ { 3 } - 24 x ^ { 2 } + 16 x = 2 x \left( 27 x ^ { 3 } - 18 x ^ { 2 } - 12 x + 8 \right)\)

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

\(2 x ( 3 x - 2 ) ^ { 2 } ( 3 x + 2 )\). The check is left to the reader.

Example \(\PageIndex{2}\):

Factor: \(x ^ { 4 } - 3 x ^ { 2 } - 4\).

This trinomial does not have a GCF.

\(\begin{aligned} x ^ { 4 } - 3 x ^ { 2 } - 4 & = \left( x ^ { 2 } \quad\right) \left( x ^ { 2 }\quad \right) \\ & = \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } - 4 \right) \quad\color{Cerulean} { Difference\: of\: squares } \\ & = \left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 ) \end{aligned}\)

The factor \(\left( x ^ { 2 } + 1 \right)\) is prime and the trinomial is completlely factored.

\(\left( x ^ { 2 } + 1 \right) ( x + 2 ) ( x - 2 )\)

Example \(\PageIndex{3}\):

Factor: \(x ^ { 6 } + 6 x ^ { 3 } - 16\).

Begin by factoring \(x ^ { 6 } = x ^ { 3 } \cdot x ^ { 3 }\) and look for the factors of \(16\) that add to \(6\).

\(\begin{aligned} x ^ { 6 } + 6 x ^ { 3 } - 16 & = \left( x ^ { 3 }\quad \right) \left( x ^ { 3 }\quad \right) \\ & = \left( x ^ { 3 } - 2 \right) \left( x ^ { 3 } + 8 \right)\quad \color{Cerulean} { sum\:of\:cubes } \\ & = \left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } - 2 x + 4 \right) \end{aligned}\)

The factor \(\left( x ^ { 3 } - 2 \right)\) cannot be factored any further using integers and the factorization is complete.

\(\left( x ^ { 3 } - 2 \right) ( x + 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

Exercise \(\PageIndex{1}\)

Factor: \(9 x ^ { 4 } + 17 x ^ { 2 } - 2\)

\(( 3 x + 1 ) ( 3 x - 1 ) \left( x ^ { 2 } + 2 \right)\)

www.youtube.com/v/lsAP8_UgUx0

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property 20 :

\(a⋅b=0\)   if and only if   \(a=0\) or \(b=0\)

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Example \(\PageIndex{4}\)

Solve: \(2 x ( x - 4 ) ( 5 x + 3 ) = 0\).

Set each variable factor equal to zero and solve.

\(\begin{aligned} 2 x & = 0 \\ \frac { 2 x } { \color{Cerulean}{2} } & = \frac { 0 } {\color{Cerulean}{ 2} } \\ x & = 0 \end{aligned}\) or \(\begin{aligned} x - 4 & = 0 \\ x & = 4 \end{aligned}\) or \(\begin{aligned} 5 x + 3 & = 0 \\ \frac { 5 x } { \color{Cerulean}{5} } & = \frac { - 3 } {\color{Cerulean}{ 5} } \\ x & = - \frac { 3 } { 5 } \end{aligned}\)

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

The solutions are \(0, 4\), and \(\frac{−3}{5}\).

Of course, most equations will not be given in factored form.

Example \(\PageIndex{5}\)

Solve: \(4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0\).

Begin by factoring the left side completely.

\(\begin{aligned} 4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 & = 0 \quad\color{Cerulean}{Factor\:by\:grouping.} \\ x ^ { 2 } ( 4 x - 1 ) - 25 ( 4 x - 1 ) & = 0 \\ ( 4 x - 1 ) \left( x ^ { 2 } - 25 \right) & = 0\quad\color{Cerulean}{Factor\:as\:a\:difference\:of\:squares.} \\ ( 4 x - 1 ) ( x + 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Set each factor equal to zero and solve.

\(\begin{array} { r } { 4 x - 1 = 0 } \\ { 4 x = 1 } \\ { x = \frac { 1 } { 4 } } \end{array}\) or \(\begin{aligned} x + 5 & = 0 \\ x & = - 5 \end{aligned}\) or \(\begin{aligned} x - 5 & = 0 \\ x & = 5 \end{aligned}\)

The solutions are \(\frac{1}{4}, −5\), and \(5\).

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring 21 are outlined in the following example.

Example \(\PageIndex{6}\)

Solve: \(15 x ^ { 2 } + 3 x - 8 = 5 x - 7\).

Step 1: Express the equation in standard form, equal to zero. In this example, subtract \(5x\) from and add \(7\) to both sides.

\(\begin{array} { l } { 15 x ^ { 2 } + 3 x - 8 = 5 x - 7 } \\ { 15 x ^ { 2 } - 2 x - 1 = 0 } \end{array}\)

Step 2: Factor the expression.

\((3x−1)(5x+1)=0\)

Step 3: Apply the zero-product property and set each variable factor equal to zero.

\(3x−1=0\)        or        \(5x+1=0\)

Step 4: Solve the resulting linear equations.

\(\begin{array} { r } { 3 x - 1 = 0 } \\ { 3 x = 1 } \\ { x = \frac { 1 } { 3 } } \end{array}\) or \(\begin{aligned} 5 x + 1 & = 0 \\ 5 x & = - 1 \\ x & = - \frac { 1 } { 5 } \end{aligned}\)

The solutions are \(\frac{1}{3}\) and \(\frac{−1}{5}\). The check is optional.

Example \(\PageIndex{7}\):

Solve: \((3x+2)(x+1)=4\).

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(4\). However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

\(\begin{array} { r } { ( 3 x + 2 ) ( x + 1 ) = 4 } \\ { 3 x ^ { 2 } + 3 x + 2 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x + 2 = 4 } \\ { 3 x ^ { 2 } + 5 x - 2 = 0 } \end{array}\)

Once it is in standard form, we can factor and then set each factor equal to zero.

        \(\begin{array} { c } { ( 3 x - 1 ) ( x + 2 ) = 0 } \\ { 3 x - 1 = 0 \quad \text { or } \quad x + 2 = 0 } \\ \quad\quad{ 3 x = 1 }\quad\quad\quad\:\:\quad x=-2\\ { x = \frac { 1 } { 3 } }\quad\quad\quad\quad\quad\: \end{array}\)

The solutions are \(\frac{1}{3}\) and \(−2\).

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

\(f ( x ) = a _ { n } x ^ { n } + a _ { n - 1 } x ^ { n - 1 } + \cdots + a _ { 1 } x + a _ { 0 }\)

A root 22 of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, \(f(x)=0\).

Example \(\PageIndex{8}\)

Find the roots: \(f ( x ) = ( x + 2 ) ^ { 2 } - 4\).

To find roots we set the function equal to zero and solve.

\(\begin{aligned} f ( x ) & = 0 \\ ( x + 2 ) ^ { 2 } - 4 & = 0 \\ x ^ { 2 } + 4 x + 4 - 4 & = 0 \\ x ^ { 2 } + 4 x & = 0 \\ x ( x + 4 ) & = 0 \end{aligned}\)

Next, set each factor equal to zero and solve.

\(\begin{aligned} x = 0 \quad \text { or } \quad x + 4 = 0 \\ x = - 4 \end{aligned}\)

We can show that these \(x\)-values are roots by evaluating.

\(\begin{aligned} f ( 0 ) & = ( 0 + 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)   \(\begin{aligned} f ( - 4 ) & = ( - 4 + 2 ) ^ { 2 } - 4 \\ & = ( - 2 ) ^ { 2 } - 4 \\ & = 4 - 4 \\ & = 0 \color{Cerulean}{✓}\end{aligned}\)

The roots are \(0\) and \(−4\).

If we graph the function in the previous example we will see that the roots correspond to the \(x\)-intercepts of the function. Here the function \(f\) is a basic parabola shifted \(2\) units to the left and \(4\) units down.

Example \(\PageIndex{9}\)

Find the roots: \(f ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4\).

\(\begin{aligned} f ( x ) & = 0 \\ x ^ { 4 } - 5 x ^ { 2 } + 4 & = 0 \\ \left( x ^ { 2 } - 1 \right) \left( x ^ { 2 } - 4 \right) & = 0 \\ ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \end{aligned}\)

\(\begin{aligned} x + 1 & = 0 \\ x & = - 1 \end{aligned}\) or \(\begin{array} { r } { x - 1 = 0 } \\ { x = 1 } \end{array}\) or \(\begin{aligned} x + 2 & = 0 \\ x & = - 2 \end{aligned}\) or \(\begin{aligned} x - 2 & = 0 \\ x & = 2 \end{aligned}\)

The roots are \(−1, 1, −2\), and \(2\).

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

Notice that the degree of the polynomial is \(4\) and we obtained four roots. In general, for any polynomial function with one variable of degree \(n\), the fundamental theorem of algebra 23 guarantees \(n\) real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Example \(\PageIndex{10}\)

Find the roots: \(f ( x ) = - x ^ { 2 } + 10 x - 25\).

\(\begin{aligned} f ( x ) & = 0 \\ - x ^ { 2 } + 10 x - 25 & = 0 \\ - \left( x ^ { 2 } - 10 x + 25 \right) & = 0 \\ - ( x - 5 ) ( x - 5 ) & = 0 \end{aligned}\)

Next, set each variable factor equal to zero and solve.

\(\begin{aligned} x - 5 & = 0 \quad\quad\text { or }& x - 5 = 0 \\ & = 5 \quad &x= 5 \end{aligned}\)

A solution that is repeated twice is called a double root 24 . In this case, there is only one solution.

The root is \(5\).

The previous example shows that a function of degree \(2\) can have one root. From the factoring step, we see that the function can be written

\(f ( x ) = - ( x - 5 ) ^ { 2 }\)

In this form, we can see a reflection about the \(x\)-axis and a shift to the right \(5\) units. The vertex is the \(x\)-intercept, illustrating the fact that there is only one root.

Exercise \(\PageIndex{2}\)

Find the roots of \(f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - x - 3\).

\(±1, −3\)

www.youtube.com/v/t1ShqqhoaVE

Example \(\PageIndex{11}\)

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\), where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(40\) feet.

We are asked to find the speed \(x\) where the safe stopping distance \(d(x)=40\) feet.

\(\begin{array} { c } { d ( x ) = 40 } \\ { \frac { 1 } { 20 } x ^ { 2 } + x = 40 } \end{array}\)

To solve for \(x\), rewrite the resulting equation in standard form. In this case, we will first multiply both sides by \(20\) to clear the fraction.

\(\begin{aligned} \color{Cerulean}{20}\color{black}{ \left( \frac { 1 } { 20 } x ^ { 2 } + x \right)} & = \color{Cerulean}{20}\color{black}{ (} 40 ) \\ x ^ { 2 } + 20 x & = 800 \\ x ^ { 2 } + 20 x - 800 & = 0 \end{aligned}\)

Next factor and then set each factor equal to zero.

\(\begin{aligned} x ^ { 2 } + 20 x - 800 & = 0 \\ ( x + 40 ) ( x - 20 ) & = 0 \\ x + 40 & = 0\quad or\quad \:x-20=0 \\ x & = - 40\quad \quad\quad\quad x=20 \end{aligned}\)

The negative answer does not make sense in the context of this problem. Consider \(x=20\) miles per hour to be the only solution.

\(20\) miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Example \(\PageIndex{12}\)

Find a quadratic equation with integer coefficients, given solutions \(\frac{−3}{2}\) and \(\frac{1}{3}\).

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

\(\begin{aligned} x & = - \frac { 3 } { 2 } \\ 2 x & = - 3 \\ 2 x + 3 & = 0 \end{aligned}\) or \(\begin{aligned} x & = \frac { 1 } { 3 } \\ 3 x & = 1 \\ 3 x - 1 & = 0 \end{aligned}\)

The product of these linear factors is equal to zero when \(x=\frac{−3}{2}\) or \(x=\frac{1}{3}\).

\((2x+3)(3x−1)=0\)

Multiply the binomials and present the equation in standard form.

\(\begin{array} { r } { 6 x ^ { 2 } - 2 x + 9 x - 3 = 0 } \\ { 6 x ^ { 2 } + 7 x - 3 = 0 } \end{array}\)

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

\(6 x ^ { 2 } + 7 x - 3 = 0\)

Example \(\PageIndex{13}\)

Find a polynomial function with real roots \(1, −2\), and \(2\).

Given solutions to \(f(x)=0\) we can find linear factors.

\(\begin{array} { r } { x = 1 } \\ { x - 1 = 0 } \end{array}\) or \(\begin{aligned} x & = - 2 \\ x + 2 & = 0 \end{aligned}\) or \(\begin{aligned} x & = 2 \\ x - 2 & = 0 \end{aligned}\)

Apply the zero-product property and multiply.

\(\begin{aligned} ( x - 1 ) ( x + 2 ) ( x - 2 ) & = 0 \\ ( x - 1 ) \left( x ^ { 2 } - 4 \right) & = 0 \\ x ^ { 3 } - 4 x - x ^ { 2 } + 4 & = 0 \\ x ^ { 3 } - x ^ { 2 } - 4 x + 4 & = 0 \end{aligned}\)

\(f ( x ) = x ^ { 3 } - x ^ { 2 } - 4 x + 4\)

Exercise \(\PageIndex{3}\)

Find a polynomial equation with integer coefficients, given solutions \(\frac{1}{2}\) and \(\frac{−3}{4}\).

\(8 x ^ { 2 } + 2 x - 3 = 0\)

www.youtube.com/v/o4cuUWWEGdU

Key Takeaways

• Factoring and the zero-product property allow us to solve equations.
• To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
• Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
• A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
• To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

Exercise \(\PageIndex{4}\)

Factor completely.

• \(50 x ^ { 2 } - 18\)
• \(12 x ^ { 3 } - 3 x\)
• \(10 x ^ { 3 } + 65 x ^ { 2 } - 35 x\)
• \(15 x ^ { 4 } + 7 x ^ { 3 } - 4 x ^ { 2 }\)
• \(6 a ^ { 4 } b - 15 a ^ { 3 } b ^ { 2 } - 9 a ^ { 2 } b ^ { 3 }\)
• \(8 a ^ { 3 } b - 44 a ^ { 2 } b ^ { 2 } + 20 a b ^ { 3 }\)
• \(36 x ^ { 4 } - 72 x ^ { 3 } - 4 x ^ { 2 } + 8 x\)
• \(20 x ^ { 4 } + 60 x ^ { 3 } - 5 x ^ { 2 } - 15 x\)
• \(3 x ^ { 5 } + 2 x ^ { 4 } - 12 x ^ { 3 } - 8 x ^ { 2 }\)
• \(10 x ^ { 5 } - 4 x ^ { 4 } - 90 x ^ { 3 } + 36 x ^ { 2 }\)
• \(x ^ { 4 } - 23 x ^ { 2 } - 50\)
• \(2 x ^ { 4 } - 31 x ^ { 2 } - 16\)
• \(- 2 x ^ { 5 } - 6 x ^ { 3 } + 8 x\)
• \(- 36 x ^ { 5 } + 69 x ^ { 3 } + 27 x\)
• \(54 x ^ { 5 } - 78 x ^ { 3 } + 24 x\)
• \(4 x ^ { 6 } - 65 x ^ { 4 } + 16 x ^ { 2 }\)
• \(x ^ { 6 } - 7 x ^ { 3 } - 8\)
• \(x ^ { 6 } - 25 x ^ { 3 } - 54\)
• \(3 x ^ { 6 } + 4 x ^ { 3 } + 1\)
• \(27 x ^ { 6 } - 28 x ^ { 3 } + 1\)

1. \(2 ( 5 x + 3 ) ( 5 x - 3 )\)

3. \(5 x ( x + 7 ) ( 2 x - 1 )\)

5. \(3 a ^ { 2 } b ( 2 a + b ) ( a - 3 b )\)

7. \(4 x ( x - 2 ) ( 3 x + 1 ) ( 3 x - 1 )\)

9. \(x ^ { 2 } ( 3 x + 2 ) ( x + 2 ) ( x - 2 )\)

11. \(\left( x ^ { 2 } + 2 \right) ( x + 5 ) ( x - 5 )\)

13. \(- 2 x \left( x ^ { 2 } + 4 \right) ( x - 1 ) ( x + 1 )\)

15. \(6 x ( x + 1 ) ( x - 1 ) ( 3 x + 2 ) ( 3 x - 2 )\)

17. \(( x + 1 ) \left( x ^ { 2 } - x + 1 \right) ( x - 2 ) \left( x ^ { 2 } + 2 x + 4 \right)\)

19. \(\left( 3 x ^ { 3 } + 1 \right) ( x + 1 ) \left( x ^ { 2 } - x + 1 \right)\)

Exercise \(\PageIndex{5}\)

• \(( 6 x - 5 ) ( x + 7 ) = 0\)
• \(( x + 9 ) ( 3 x - 8 ) = 0\)
• \(5 x ( 2 x - 5 ) ( 3 x + 1 ) = 0\)
• \(4 x ( 5 x - 1 ) ( 2 x + 3 ) = 0\)
• \(( x - 1 ) ( 2 x + 1 ) ( 3 x - 5 ) = 0\)
• \(( x + 6 ) ( 5 x - 2 ) ( 2 x + 9 ) = 0\)
• \(( x + 4 ) ( x - 2 ) = 16\)
• \(( x + 1 ) ( x - 7 ) = 9\)
• \(( 6 x + 1 ) ( x + 1 ) = 6\)
• \(( 2 x - 1 ) ( x - 4 ) = 39\)
• \(x ^ { 2 } - 15 x + 50 = 0\)
• \(x ^ { 2 } + 10 x - 24 = 0\)
• \(3 x ^ { 2 } + 2 x - 5 = 0\)
• \(2 x ^ { 2 } + 9 x + 7 = 0\)
• \(\frac { 1 } { 10 } x ^ { 2 } - \frac { 7 } { 15 } x - \frac { 1 } { 6 } = 0\)
• \(\frac { 1 } { 4 } - \frac { 4 } { 9 } x ^ { 2 } = 0\)
• \(6 x ^ { 2 } - 5 x - 2 = 30 x + 4\)
• \(6 x ^ { 2 } - 9 x + 15 = 20 x - 13\)
• \(5 x ^ { 2 } - 23 x + 12 = 4 ( 5 x - 3 )\)
• \(4 x ^ { 2 } + 5 x - 5 = 15 ( 3 - 2 x )\)
• \(( x + 6 ) ( x - 10 ) = 4 ( x - 18 )\)
• \(( x + 4 ) ( x - 6 ) = 2 ( x + 4 )\)
• \(4 x ^ { 3 } - 14 x ^ { 2 } - 30 x = 0\)
• \(9 x ^ { 3 } + 48 x ^ { 2 } - 36 x = 0\)
• \(\frac { 1 } { 3 } x ^ { 3 } - \frac { 3 } { 4 } x = 0\)
• \(\frac { 1 } { 2 } x ^ { 3 } - \frac { 1 } { 50 } x = 0\)
• \(- 10 x ^ { 3 } - 28 x ^ { 2 } + 48 x = 0\)
• \(- 2 x ^ { 3 } + 15 x ^ { 2 } + 50 x = 0\)
• \(2 x ^ { 3 } - x ^ { 2 } - 72 x + 36 = 0\)
• \(4 x ^ { 3 } - 32 x ^ { 2 } - 9 x + 72 = 0\)
• \(45 x ^ { 3 } - 9 x ^ { 2 } - 5 x + 1 = 0\)
• \(x ^ { 3 } - 3 x ^ { 2 } - x + 3 = 0\)
• \(x ^ { 4 } - 5 x ^ { 2 } + 4 = 0\)
• \(4 x ^ { 4 } - 37 x ^ { 2 } + 9 = 0\)

1. \(- 7 , \frac { 5 } { 6 }\)

3. \(0 , \frac { 5 } { 2 } , - \frac { 1 } { 3 }\)

5. \(- \frac { 1 } { 2 } , 1 , \frac { 5 } { 3 }\)

7. \(- 6,4\)

9. \(- \frac { 5 } { 3 } , \frac { 1 } { 2 }\)

11. \(5,10\)

13. \(- \frac { 5 } { 3 } , 1\)

15. \(- \frac { 1 } { 3 } , 5\)

17. \(- \frac { 1 } { 6 } , 6\)

19. \(\frac { 3 } { 5 } , 8\)

21. \(2,6\)

23. \(0 , - \frac { 3 } { 2 } , 5\)

25. \(0 , \pm \frac { 3 } { 2 }\)

27. \(- 4,0 , \frac { 6 } { 5 }\)

29. \(\pm 6 , \frac { 1 } { 2 }\)

31. \(\pm \frac { 1 } { 3 } , \frac { 1 } { 5 }\)

33. \(\pm 1 , \pm 2\)

Exercise \(\PageIndex{6}\)

Find the roots of the given functions.

• \(f ( x ) = x ^ { 2 } + 10 x - 24\)
• \(f ( x ) = x ^ { 2 } - 14 x + 48\)
• \(f ( x ) = - 2 x ^ { 2 } + 7 x + 4\)
• \(f ( x ) = - 3 x ^ { 2 } + 14 x + 5\)
• \(f ( x ) = 16 x ^ { 2 } - 40 x + 25\)
• \(f ( x ) = 9 x ^ { 2 } - 12 x + 4\)
• \(g ( x ) = 8 x ^ { 2 } + 3 x\)
• \(g ( x ) = 5 x ^ { 2 } - 30 x\)
• \(p ( x ) = 64 x ^ { 2 } - 1\)
• \(q ( x ) = 4 x ^ { 2 } - 121\)
• \(f ( x ) = \frac { 1 } { 5 } x ^ { 3 } - 1 x ^ { 2 } - \frac { 1 } { 20 } x + \frac { 1 } { 4 }\)
• \(f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } - \frac { 4 } { 3 } x - 2\)
• \(g ( x ) = x ^ { 4 } - 13 x ^ { 2 } + 36\)
• \(g ( x ) = 4 x ^ { 4 } - 13 x ^ { 2 } + 9\)
• \(f ( x ) = ( x + 5 ) ^ { 2 } - 1\)
• \(g ( x ) = - ( x + 5 ) ^ { 2 } + 9\)
• \(f ( x ) = - ( 3 x - 5 ) ^ { 2 }\)
• \(g ( x ) = - ( x + 2 ) ^ { 2 } + 4\)

1. \(2 , - 12\)

3. \(- \frac { 1 } { 2 } , 4\)

5. \(\frac { 5 } { 4 }\)

7. \(- \frac { 3 } { 8 } , 0\)

9. \(\pm \frac { 1 } { 8 }\)

11. \(\pm \frac { 1 } { 2 } , 5\)

13. \(\pm 2 , \pm 3\)

15. \(- 6 , - 4\)

17. \(\frac { 5 } { 3 }\)

Exercise \(\PageIndex{7}\)

Given the graph of a function, determine the real roots.

5. The sides of a square measure \(x − 2\) units. If the area is \(36\) square units, then find \(x\).

6. The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

7. The profit in dollars generated by producing and selling n bicycles per week is given by the formula \(P ( n ) = - 5 n ^ { 2 } + 400 n - 6000\). How many bicycles must be produced and sold to break even?

8. The height in feet of an object dropped from the top of a \(64\)-foot building is given by \(h ( t ) = - 16 t ^ { 2 } + 64\) where \(t\) represents the time in seconds after it is dropped. How long will it take to hit the ground?

9. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

What is the length of each side of the cardboard sheet if the volume of the box is to be \(98\) cubic inches?

10. The height of a triangle is \(4\) centimeters less than twice the length of its base. If the total area of the triangle is \(48\) square centimeters, then find the lengths of the base and height.

11. A uniform border is to be placed around an \(8 × 10\) inch picture.

If the total area including the border must be \(168\) square inches, then how wide should the border be?

12. The area of a picture frame including a \(3\)-inch wide border is \(120\) square inches.

If the width of the inner area is \(2\) inches less than its length, then find the dimensions of the inner area.

13. Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by \(d ( x ) = \frac { 1 } { 20 } x ^ { 2 } + x\) where \(x\) represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in \(75\) feet.

14. A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula \(R ( n ) = 12 n - 0.6 n ^ { 2 }\) where \(n\) represents the number of palettes of product sold \((0 ≤ n < 20)\). Determine the number of palettes sold in a day if the revenue was \(45\) thousand dollars.

1. \(- 3 , - 1,0,2\)

3. \(- 2,3\)

5. \(8\) units

7. \(20\) or \(60\) bicycles

9. \(11\) in

11. \(2\) inches

13. \(30\) miles per hour

Exercise \(\PageIndex{8}\)

Find a polynomial equation with the given solutions.

• \(2, \frac{1}{3}\)
• \(- \frac { 3 } { 4 } , 5\)
• \(- 3,1,3\)
• \(- 5 , - 1,1\)

1. \(x ^ { 2 } - 2 x - 15 = 0\)

3. \(3 x ^ { 2 } - 7 x + 2 = 0\)

5. \(x ^ { 2 } + 4 x = 0\)

7. \(x ^ { 2 } - 49 = 0\)

9. \(x ^ { 3 } - x ^ { 2 } - 9 x + 9 = 0\)

Exercise \(\PageIndex{9}\)

Find a function with the given roots.

• \(\frac { 1 } { 2 } , \frac { 2 } { 3 }\)
• \(\frac { 2 } { 5 } , - \frac { 1 } { 3 }\)
• \(\pm \frac { 3 } { 4 }\)
• \(\pm \frac { 5 } { 2 }\)
• \(5\) double root
• \(-3\) double root

1. \(f ( x ) = 6 x ^ { 2 } - 7 x + 2\)

3. \(f ( x ) = 16 x ^ { 2 } - 9\)

5. \(f ( x ) = x ^ { 2 } - 10 x + 25\)

7. \(f ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 3 x\)

Exercise \(\PageIndex{10}\)

Recall that if \(| X | = p\), then \(X=-p\) or \(X=p\). Use this to solve the following absolute value equations.

• \(\left| x ^ { 2 } - 8 \right| = 8\)
• \(\left| 2 x ^ { 2 } - 9 \right| = 9\)
• \(\left| x ^ { 2 } - 2 x - 1 \right| = 2\)
• \(\left| x ^ { 2 } - 8 x + 14 \right| = 2\)
• \(\left| 2 x ^ { 2 } - 4 x - 7 \right| = 9\)
• \(\left| x ^ { 2 } - 3 x - 9 \right| = 9\)

1. \(\pm 4,0\)

3. \(\pm 1,3\)

5. \(- 2,1,4\)

Exercise \(\PageIndex{11}\)

• Explain to a beginning algebra student the difference between an equation and an expression.
• What is the difference between a root and an \(x\)-intercept? Explain.
• Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.
• Research and discuss the fundamental theorem of algebra.

1. Answer may vary

3. Answer may vary

20 A product is equal to zero if and only if at least one of the factors is zero.

21 The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero.

22 A value in the domain of a function that results in zero.

23 Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree.

24 A root that is repeated twice.

Zeros of Polynomial Functions

Solve real-world applications of polynomial equations.

We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.

Example 8: Solving Polynomial Equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\frac{1}{3}w[/latex]. Let’s write the volume of the cake in terms of width of the cake.

Substitute the given volume into this equation.

Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351[/latex], and [latex]\pm 1053[/latex]. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[/latex].

Since 1 is not a solution, we will check [latex]x=3[/latex].

Since 3 is not a solution either, we will test [latex]x=9[/latex].

Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.

The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?

3 meters by 4 meters by 7 meters

• Precalculus. Authored by : Jay Abramson, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download For Free at : http://cnx.org/contents/[email protected].

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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