Three Techniques for Evaluating and Finding Zeros of Polynomial Functions

In this section, you will:

• Evaluate a polynomial using the Remainder Theorem.
• Use the Factor Theorem to solve a polynomial equation.
• Use the Rational Zero Theorem to find rational zeros.
• Find zeros of a polynomial function.
• Use the Linear Factorization Theorem to find polynomials with given zeros.
• Use Descartes' Rule of Signs.
• Solve real-world applications of 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?

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.

Source: Rice University, https://openstax.org/books/college-algebra/pages/5-5-zeros-of-polynomial-functions
This work is licensed under a Creative Commons Attribution 4.0 License.

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 $d(x)$, there exist unique polynomials $q(x)$ and $r(x)$ such that

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

and either $r(x)=0$ or the degree of $r(x)$ is less than the degree of $d(x)$. In practice divisors, $d(x)$ will have degrees less than or equal to the degree of $f(x)$. If the divisor, $d(x)$, is $x−k$, this takes the form

$f(x)=(x-k) q(x)+r$

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

$\begin{aligned} f(k) &=(k-k) q(k)+r \\ &=0 \cdot q(k)+r \\ &=r \end{aligned}$

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 $f(k)$.

HOW TO

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 1

Using the Remainder Theorem to Evaluate a Polynomial

Use the Remainder Theorem to evaluate $f(x)=6 x^{4}-x^{3}-15 x^{2}+2 x-7 \text { at } x=2 .$.

Solution

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

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

Analysis

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

$\begin{aligned} f(x) &=6 x^{4}-x^{3}-15 x^{2}+2 x-7 \\ f(2) &=6(2)^{4}-(2)^{3}-15(2)^{2}+2(2)-7 \\ &=25 \end{aligned}$

TRY IT #1

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

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.

EXAMPLE 2

Using the Factor Theorem to Find the Zeros of a Polynomial Expression

Show that $(x+2)$ is a factor of $x^{3}-6 x^{2}-x+30$. 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.

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:

$(x+2)\left(x^{2}-8 x+15\right)$

We can factor the quadratic factor to write the polynomial as

$(x+2)(x−3)(x−5)$

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

TRY IT #2

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

Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial.

Consider a quadratic function with two zeros, $x=\frac{2}{5}$ and $x=\frac{3}{4}$. By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor.

$\begin{array}{ll}x-\frac{2}{5}=0 \text { or } x-\frac{3}{4}=0 & \text { Set each factor equal to } 0 . \\5 x-2=0 \text { or } 4 x-3=0 & \text { Multiply both sides of the equation to eliminate fractions. } \\f(x)=(5 x-2)(4 x-3) & \text { Create the quadratic function, multiplying the factors. } \\f(x)=20 x^{2}-23 x+6 & \text { Expand the polynomial. } \\f(x)=(5 \cdot 4) x^{2}-23 x+(2 \cdot 3) &\end{array}$

Notice that two of the factors of the constant term, $6$, are the two numerators from the original rational roots: $2$ and $3$. Similarly, two of the factors from the leading coefficient, $20$, are the two denominators from the original rational roots: $5$ and $4$.

We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros.

THE RATIONAL ZERO THEOREM

The Rational Zero Theorem states that, if the polynomial $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$ has integer coefficients, then every rational zero of $f(x)$ has the form $\frac{p}{q}$ where $p$ is a factor of the constant term $a_{0}$ and $q$ is a factor of the leading coefficient $a_{n}$.

When the leading coefficient is $1$, the possible rational zeros are the factors of the constant term.

HOW TO

Given a polynomial function $f(x)$, use the Rational Zero Theorem to find rational zeros.

EXAMPLE 3

Listing All Possible Rational Zeros

List all possible rational zeros of $f(x)=2 x^{4}-5 x^{3}+x^{2}-4$.

Solution

The only possible rational zeros of $f(x)$ are the quotients of the factors of the last term, $–4$, and the factors of the leading coefficient, $2$.

The constant term is $–4$; the factors of $–4$ are $p=\pm 1, \pm 2, \pm 4$.

The leading coefficient is $2$; the factors of $2$ are $q=\pm 1, \pm 2$.

If any of the four real zeros are rational zeros, then they will be of one of the following factors of $–4$ divided by one of the factors of $2$.

$\frac{p}{q}=\pm \frac{1}{1}, \quad \pm \frac{1}{2} \quad \frac{p}{q}=\pm \frac{2}{1}, \pm \quad \frac{2}{2} \quad \frac{p}{q}=\quad \pm \quad \frac{4}{1}, \pm \quad \frac{4}{2}$

Note that $\frac{2}{2}=1$ and $\frac{4}{2}=2$, which have already been listed. So we can shorten our list.

$\frac{p}{q}=\frac{\text { Factors of the last }}{\text { Factors of the first }}=\pm 1, \pm 2, \pm 4, \pm \frac{1}{2}$

EXAMPLE 4

Using the Rational Zero Theorem to Find Rational Zeros

Use the Rational Zero Theorem to find the rational zeros of $f(x)=2 x^{3}+x^{2}-4 x+1$.

Solution

The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f(x)$, then $p$ is a factor of $1$ and $q$ is a factor of $2$.

$\begin{aligned} \frac{p}{q} &=\frac{\text { factor of constant term }}{\text { factor of leading coefficient }} \\ &=\frac{\text { factor of } 1}{\text { factor of } 2} \end{aligned}$

The factors of $1$ are $\pm 1$ and the factors of $2$ are $\pm 1$ and $\pm 2$. The possible values for $\frac{p}{q}$ are $\pm 1$ and $\pm \frac{1}{2}$. These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for $x$ in $f(x)$.

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

Of those, $−1$, $-\frac{1}{2}$,and $\frac {1}{2}$ are not zeros of $f(x)$. $1$ is the only rational zero of $f(x)$.

TRY IT #3

Use the Rational Zero Theorem to find the rational zeros of $f(x)=x^{3}-5 x^{2}+2 x+1$.

The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.

HOW TO

Given a polynomial function $f$, use synthetic division to find its zeros.

EXAMPLE 5

Finding the Zeros of a Polynomial Function with Repeated Real Zeros

Find the zeros of $f(x)=4 x^{3}-3 x-1$.

Solution

The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f(x)$, then $p$ is a factor of $–1$ and $q$ is a factor of $4$.

$\begin{aligned} \frac{p}{q} &=\frac{\text { factor of constant term }}{\text { factor of leading coefficient }} \\ &=\frac{\text { factor of }-1}{\text { factor of } 4} \end{aligned}$

The factors of $–1$ are ±1 and the factors of $4$ are $\pm 1, \pm 2$, and $\pm 4$. The possible values for $\frac{p}{q}$ are $\pm 1, \pm \frac{1}{2}$, and $\pm \frac{1}{4}$. These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of $0$. Let's begin with $1$.

Dividing by $(x−1)$ gives a remainder of $0$, so $1$ is a zero of the function. The polynomial can be written as

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

The quadratic is a perfect square. $f(x)$ can be written as

$(x-1)(2 x+1)^{2}$

We already know that $1$ is a zero. The other zero will have a multiplicity of $2$ because the factor is squared. To find the other zero, we can set the factor equal to $0$.

$\begin{aligned} 2 x+1 &=0 \\ x &=-\frac{1}{2} \end{aligned}$

The zeros of the function are $1$ and $-\frac{1}{2}$ with multiplicity $2$.

Analysis

Look at the graph of the function $f$ in Figure 1. Notice, at $x=−0.5$, the graph bounces off the $x$-axis, indicating the even multiplicity $(2,4,6…)$ for the zero $−0.5$. At $x=1$, the graph crosses the $x$-axis, indicating the odd multiplicity $(1,3,5…)$ for the zero $x=1$.

Figure 1