# 1.2: Zeros of Polynomial Functions


Learning Objectives

• Evaluate a polynomial using the Remainder Theorem.
• Use the Factor Theorem to solve a polynomial equation.

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?

This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.

## Evaluating a Polynomial Using the Remainder Theorem

In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by $$x–k$$, the remainder may be found quickly by evaluating the polynomial function at $$k$$, that is, $$f(k)$$. Let’s walk through the proof of the theorem.

Recall that the Division Algorithm states that, given a polynomial dividend $$f(x)$$ and a non-zero polynomial divisor $$d(x)$$ where the degree of $$d(x)$$ is less than or equal to the degree of $$f(x)$$,there exist unique polynomials $$q(x)$$ and $$r(x)$$ such that

$f(x)=d(x)q(x)+r(x) \nonumber$

If the divisor, $$d(x)$$, is $$x−k$$, this takes the form

$f(x)=(x−k)q(x)+r \nonumber$

Since the divisor $$x−k$$

is linear, the remainder will be a constant, $$r$$. And, if we evaluate this for $$x=k$$, we have

\begin{align*} f(k)&=(k−k)q(k)+r \\[4pt] &=0{\cdot}q(k)+r \\[4pt] &=r \end{align*}

In other words, $$f(k)$$ is the remainder obtained by dividing $$f(x)$$by $$x−k$$.

The Remainder Theorem

If a polynomial $$f(x)$$ is divided by $$x−k$$,then the remainder is the value $$f(k)$$.

Given a polynomial function $$f$$, evaluate $$f(x)$$ at $$x=k$$ using the Remainder Theorem.

1. Use synthetic division to divide the polynomial by $$x−k$$.
2. The remainder is the value $$f(k)$$.

Example $$\PageIndex{1}$$: Using the Remainder Theorem to Evaluate a Polynomial

Use the Remainder Theorem to evaluate $$f(x)=6x^4−x^3−15x^2+2x−7$$ at $$x=2$$.

Solution

To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by $$x−2$$.

$2\begin{array}{|ccccc} \; 6 & −1 & −15 & 2 & −7 \\ \text{} & 12 & 22 & 14 & 32 \\ \hline \end{array} \\ \begin{array}{ccccc} \hspace{.3cm} 6 & 11 & \; 7 & \;\;16 & \;\; 25 \end{array}$

The remainder is 25. Therefore, $$f(2)=25$$.

Analysis

We can check our answer by evaluating $$f(2)$$.

\begin{align*} f(x)&=6x^4−x^3−15x^2+2x−7 \\ f(2)&=6(2)^4−(2)^3−15(2)^2+2(2)−7 \\ &=25 \end{align*}

Exercise $$\PageIndex{1}$$

Use the Remainder Theorem to evaluate $$f(x)=2x^5−3x^4−9x^3+8x^2+2$$ at $$x=−3$$.

$$f(−3)=−412$$

## Using the Factor Theorem to Solve a Polynomial Equation

The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm.

$f(x)=(x−k)q(x)+r$

If $$k$$ is a zero, then the remainder $$r$$ is $$f(k)=0$$ and $$f (x)=(x−k)q(x)+0$$ or $$f(x)=(x−k)q(x)$$.

Notice, written in this form, $$x−k$$ is a factor of $$f(x)$$. We can conclude if $$k$$ is a zero of $$f(x)$$, then $$x−k$$ is a factor of $$f(x)$$.

Similarly, if $$x−k$$ is a factor of $$f(x)$$, then the remainder of the Division Algorithm $$f(x)=(x−k)q(x)+r$$ is $$0$$. This tells us that $$k$$ is a zero.

This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree $$n$$ in the complex number system will have $$n$$ zeros. We can use the Factor Theorem to completely factor a polynomial into the product of $$n$$ factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.

THE FACTOR THEOREM

According to the Factor Theorem, $$k$$ is a zero of $$f(x)$$ if and only if $$(x−k)$$ is a factor of $$f(x)$$.

How to: Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial

1. Use synthetic division to divide the polynomial by $$(x−k)$$.
2. Confirm that the remainder is $$0$$.
3. Write the polynomial as the product of $$(x−k)$$ and the quadratic quotient.
4. If possible, factor the quadratic.
5. Write the polynomial as the product of factors.

Example $$\PageIndex{2}$$: Using the Factor Theorem to Solve a Polynomial Equation

Show that $$(x+2)$$ is a factor of $$x^3−6x^2−x+30$$. Find the remaining factors. Use the factors to determine the zeros of the polynomial.

Solution

We can use synthetic division to show that $$(x+2)$$ is a factor of the polynomial.

$-2 \begin{array}{|cccc} \; 1 & −6 & −1 & 30 \\ \text{} & -2 & 16 & -30 \\ \hline \end{array} \\ \begin{array}{cccc} 1 & -8 & \; 15 & \;\;0 \end{array}$

The remainder is zero, so $$(x+2)$$ is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:

$(x+2)(x^2−8x+15)$

We can factor the quadratic factor to write the polynomial as

$(x+2)(x−3)(x−5)$

By the Factor Theorem, the zeros of $$x^3−6x^2−x+30$$ are –2, 3, and 5.

Exercise $$\PageIndex{2}$$

Use the Factor Theorem to find the zeros of $$f(x)=x^3+4x^2−4x−16$$ given that $$(x−2)$$ is a factor of the polynomial.