Topics
2. Polynomial Expressions and Equations
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United States of America
Grades 9-12

2.06 Solving polynomial equations

Lesson
Practice

Introduction

We previously reviewed several different strategies for factoring polynomials, including using polynomial identities. We also learned how to identify factors of a polynomial using the factor theorem and polynomial division. In this lesson, we will see how those strategies can be applied to help us solve polynomial equations.

Solving polynomial equations

A polynomial equation is an equation with polynomial expressions on both sides of the equation.

The standard form of a polynomial equation is given by a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_2x^2+a_1x+a_0=0 where n is a positive integer and a_n,a_{n-1},a_{n-2},\ldots,a_2,a_1,a_0 are constant coefficients.

We can find the roots of this polynomial equation by factoring. When a polynomial expression is factored, we can use the zero product property that if A\cdot B \cdot C=0, then A=0, B=0, or C=0.

If a polynomial cannot be factored using a familiar method, we can use the rational roots theorem to help us find the rational roots of the polynomial.

Rational roots theorem

Given P\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_2x^2+a_1x+a_0, where the coefficients are all integers. If P\left(x\right) has any rational roots, then they must be of the form \pm \dfrac{p}{q} where p is a factor of the constant term a_0 and q is a factor of the leading coefficient a_n.

For example, consider the polynomial P\left(x\right)=2x^4-3x^3+7x-9. The coefficients are all integers, so we can use the rational roots theorem to determine the possible rational roots. The constant is -9 and its factors are 1,3,9. The leading coefficient is 2, and its factors are 1,2. By the theorem, the possible rational roots of this polynomial are \pm \dfrac{1}{1}, \pm \dfrac{1}{2}, \pm \dfrac{3}{1}, \pm \dfrac{3}{2}, \pm \dfrac{9}{1}, \pm \dfrac{9}{2}

After using the theorem to find the possible rational roots, we can use the factor theorem to find the actual roots. Recall if P\left(a\right)=0, then a is a root. Once we find a root, we can write the root as a linear factor and divide the polynomial by the linear factor. Repeating this process can help us find the complete factorization of the polynomial.

Examples

Example 1

Solve the equation \left(4x^2-81\right)\left(x^2-3x-10\right)=0.

Worked Solution
Create a strategy

First, we determine whether the expression on the left-hand side of the equation can be expressed as a product of linear factors. If so, we can apply some factoring techniques.

Apply the idea

The equation contains a factored expression on the left-hand side and 0 on the right-hand side. The factors are both of degree 2. We need to express these quadratic factors as linear factors, if possible.

Observe that 4x^2-81=\left(2x\right)^2-\left(9\right)^2. This means that 4x^2-81 is a difference of two squares, so we have 4x^2-81=\left(2x-9\right)\left(2x+9\right).

\displaystyle \left(4x^2-81\right)\left(x^2-3x-10\right)\displaystyle =\displaystyle 0Original equation
\displaystyle \left(2x-9\right)\left(2x+9\right)\left(x^2-3x-10\right)\displaystyle =\displaystyle 0Difference of two squares identity
\displaystyle \left(2x-9\right)\left(2x+9\right)\left(x-5\right)\left(x+2\right)\displaystyle =\displaystyle 0Factor the quadratic

The zero product property states that if A \cdot B \cdot C \cdot D=0, then A =0, B=0, C=0 or D=0.

So we get the following equations:\begin{aligned}2x-9=0\\ 2x+9=0\\x-5=0 \\x+2=0\end{aligned} Solving each equation for x, we obtain the following solutions: x=\dfrac{9}{2}, x=-\dfrac{9}{2},x=5,x=-2

Reflect and check

We can check the answer by using technology to graph y=\left(4x^2-81\right)\left(x^2-3x-10\right), then looking for the x-intercepts.

-5
-4
-3
-2
-1
1
2
3
4
5
x
-300
-150
150
300
450
600
750
900
y

The function is equal to zero at x=-\dfrac{9}{2},x=-2, x=\dfrac{9}{2}, x=5 which confirms these are the solutions of \left(4x^2-81\right)\left(x^2-3x-10\right)=0.

Example 2

Given the polynomial p\left(x\right)=x^4-8x^3+15x^2+4x-20.

a

Find all the possible roots.

Worked Solution
Create a strategy

By the rational roots theorem, the possible roots are found by dividing each factor of the constant by each factor of the leading coefficient.

This theorem can only be used if all the coefficients in the polynomial are integers. 1, -8, 15, 4, and -20 are all integers, so this theorem can be applied to find the possible roots.

Apply the idea

The factors of the constant are 1, 2, 4, 5, 10, 20. The leading coefficient is 1.

Dividing each factor of the constant by 1 does not change the values of the factors, so the possible roots are \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.

b

Find all the roots.

Worked Solution
Create a strategy

We can use the factor theorem to determine if the possible roots from part (a) are actual roots by seeing if p\left(a\right)=0 where a is one of the possible roots. Substituting all 12 possible roots would be tedious, so we can save time by using synthetic division.

After finding one of the actual roots, we will use synthetic division to divide the polynomial by the related factor and continue factoring from there.

Apply the idea

Let's begin by using the factor theorem with the positive possible roots to see if we find one that is an actual root. We will start with x=1:

p\left(1\right)=\left(1\right)^4-8\left(1\right)^3+15\left(1\right)^2+4\left(1\right)-20=-8

This does not equal zero, so x=1 is not a root. Next, let's try x=2:

p\left(2\right)=\left(2\right)^4-8\left(2\right)^3+15\left(2\right)^2+4\left(2\right)-20=0

This shows x=2 is a root. Its related factor is \left(x-2\right). Using synthetic division to divide \dfrac{p\left(x\right)}{x-2}, we get:

A figure showing a synthetic division with 2 in the top left corner. Top row has digits 1, negative 8, 15, 4, and negative 20. Middle row has 2 under the negative 8, 15 under the negative 12, 6 under the 4, and 20 under the negative 20. Bottom row has 1, negative 6, 3, 10, and 0.

This shows \dfrac{p\left(x\right)}{x-2}=x^3-6x^2+3x+10 or p\left(x\right)=\left(x-2\right)\left(x^3-6x^2+3x+10\right). Usually, we would check to see if the cubic can be factored, but there is not an obvious way to factor the expression. So, we need to use the rational roots theorem again to find the possible roots of the cubic.

The constant of the cubic is 10, and the leading coefficient is 1. The possbile factors are \pm 1, \pm 2, \pm 5, \pm 10. We will try substituting values into the cubic until we find one that results in zero:

\begin{aligned}\left(1\right)^3-6\left(1\right)^2+3\left(1\right)+10&=8\\ \left(2\right)^3-6\left(2\right)^2+3\left(2\right)+10&=0\end{aligned}

This means x=2 is another root of the polynomial. It is possible for a root to be repeated, as we have just shown. We will use synthetic division again to divide the cubic by x-2 to continue factoring the polynomial.

A figure showing a synthetic division with 2 in the top left corner. Top row has digits 1, negative 6, 3, and 10. Middle row has 2 under the negative 6, negative 8 under the 3, and negative 10 under the 10. Bottom row has 1, negative 4, negative 5, and 0.

The result of \dfrac{x^3-6x^2+3x+10}{x-2}=x^2-4x-5. This means the factorization of p\left(x\right) so far is \left(x-2\right)\left(x-2\right)\left(x^2-4x-5\right). From here, we can see that the resulting quadratic is factorable. Once we factor it, we can find the remaining roots.

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

Setting each of the factors equal to zero and solving, we find the roots are x=2,5,-1.

Reflect and check

We can check our answer by graphing the polynomial with technology and identifying the x-intercepts.

-2
-1
1
2
3
4
5
6
x
-20
-15
-10
-5
5
10
15
20
y

Here, we see the roots of the polynomial are x=-1, 2, 5 which confirms our answer.

Example 3

Solve the equation 4x^5-9x^3=72-32x^2.

Worked Solution
Create a strategy

We need to first determine whether the equation is in standard form. If it is not, we need to convert it into standard form before we apply some factoring techniques to obtain the solutions.

Apply the idea

We can see that the equation 4x^5-9x^3=72-32x^2 is not in standard form because it is not equal to 0. We can convert the equation to standard form by moving the terms 72 and -32x^2 to the left-hand side of the equation using inverse operations. This leaves 0 on the right-hand side of the equation. The equation in standard form is 4x^5-9x^3+32x^2-72=0

We can proceed by factoring the left-hand side by grouping

\displaystyle 4x^5-9x^3+32x^2-72\displaystyle =\displaystyle 0Standard form
\displaystyle (4x^5+32x^2)+(-9x^3-72)\displaystyle =\displaystyle 0Group terms based on common factors
\displaystyle 4x^2(x^3+8)-9(x^3+8)\displaystyle =\displaystyle 0Factor out the GCF: 4x^2 and -9
\displaystyle \left(4x^2-9\right)\left(x^3+8\right)\displaystyle =\displaystyle 0Factor out the common binomial

Observe that 4x^2-9=(2x)^2-(3)^2. This means that 4x^2-9 is a difference of two squares, so we have 4x^2-9=\left(2x-3\right)\left(2x+3\right).

\displaystyle \left(2x-3\right)\left(2x+3\right)\left(x^3+8\right)\displaystyle =\displaystyle 0Difference of two squares identity

Also, observe that x^3+8=(x)^3+(2)^3. This means that x^3+8 is a sum of two cubes, so we have x^3+8=(x+2)\left(x^2-2x+4\right).

\displaystyle \left(2x-3\right)\left(2x+3\right)\left(x+2\right)\left(x^2-2x+4\right)\displaystyle =\displaystyle 0Sum of two cubes

The zero product property states that if A \cdot B \cdot C \cdot D=0, then A =0, B=0, C=0 or D=0.

So we get the following equations:\begin{aligned}2x-3=0\\ 2x+3=0\\x+2=0\\x^2-2x+4=0\end{aligned} Solving the first three equations for x, we obtain the following solutions:

\begin{aligned}2x-3&=0, x=\frac{3}{2}\\ 2x+3&=0, x=-\frac{3}{2}\\x+2&=0, x=-2\end{aligned}

Since the expression x^2-2x+4 is not factorable, we will use the quadratic formula in solving x^2-2x+4=0:

\displaystyle x\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-(-2)\pm\sqrt{\left(-2\right)^2-4(1)(4)}}{2(1)}Substitute a=1,b=-2 and c=4
\displaystyle x\displaystyle =\displaystyle 1\pm \sqrt{3}iEvaluate the operations

Therefore, the solutions are:x=\dfrac{3}{2},x=-\dfrac{3}{2},x=-2,x=1+ \sqrt{3}i,x=1- \sqrt{3}i

Idea summary

The roots of a polynomial equation, p\left(x\right)=0, can be found by factoring. After p\left(x\right) has been factored, we can use the zero product property to find the roots.

If a polynomial cannot be factored using a familiar method or identity, we can use the rational roots theorem to identity possible roots. The theorem tells us to divide each factor of the constant by each factor of the leading coefficient to determine the positive and negative possible rational roots.

Outcomes

N.CN.C.7

Solve quadratic equations with real coefficients that have complex solutions

N.CN.C.8 (+)

Extend polynomial identities to the complex numbers.

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.B.3

Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

A.CED.A.4

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.

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