Zeros of Polynomial Functions

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, $f\left(k\right)$ 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

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

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

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

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 tells us

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.

Using the Rational Zero Theorem to Find Rational Zeros

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.

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.

Finding the Zeros of Polynomial Functions

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.

Example 5

Finding the Zeros of a Polynomial Function with Repeated Real Zeros

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

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.

$$\begin{array}{l}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ =\frac{\text{factor of –1}}{\text{factor of 4}}\hfill \end{array}$$

The factors of $–1$ are $\mathrm{\pm 1}$ and the factors of $4$ are $\mathrm{\pm 1},\mathrm{\pm 2},$ and $\mathrm{\pm 4.}$ The possible values for $\frac{p}{q}$ are $\mathrm{\pm 1},\pm \frac{1}{2},$ 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.

$$\begin{array}{l}\\ \begin{array}{c}1\\ \end{array}\overline{)\begin{array}{rrrr}\hfill 4& \hfill 0& \hfill -3& \hfill -1\\ \hfill & \hfill 4& \hfill 4& \hfill 1\end{array}}\\ \begin{array}{llll}\text{ 4}\hfill & \text{ 4}\hfill & \text{ 1}\hfill & 0\hfill \end{array}\end{array}$$

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

$$(x-1)(4x^2+4x+1).$$

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

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

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.

$$\begin{array}{l}2x+1=0\hfill \\ x=-\frac{1}{2}\hfill \end{array}$$

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 $-\mathrm{0.5.}$ At $x=1,$ the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero $x=1.$

Using the Fundamental Theorem of Algebra

Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.

Suppose $f$ is a polynomial function of degree four, and $f(x)=0.$ The Fundamental Theorem of Algebra states that there is at least one complex solution, call it $c_1.$ By the Factor Theorem, we can write $f(x)$ as a product of $x-c_{\text{1}}$ and a polynomial quotient. Since $x-c_{\text{1}}$ is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it $c_{\text{2}}.$ So we can write the polynomial quotient as a product of $x-c_{\text{2}}$ and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of $f(x).$

Example 6

Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of $f(x)=3x^3+9x^2+x+3.$

Solution

The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f(x),$ then $p$ is a factor of 3 and $q$ is a factor of 3.

$$\begin{array}{l}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ =\frac{\text{factor of 3}}{\text{factor of 3}}\hfill \end{array}$$

The factors of 3 are $\mathrm{\pm 1}$ and $\mathrm{\pm 3.}$ The possible values for $\frac{p}{q},$ and therefore the possible rational zeros for the function, are $\mathrm{\pm 3},\text{±1, and }\pm \frac{1}{3}.$ We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3.

$$\begin{array}{l}\hfill \\ \begin{array}{c}-3\\ \end{array}\overline{)\begin{array}{rrrr}\hfill 3& \hfill 9& \hfill 1& \hfill 3\\ \hfill & \hfill -9& \hfill 0& \hfill -3\end{array}}\hfill \\ \begin{array}{rrrr}\hfill \text{ 3}& \hfill 0& \hfill 1& \hfill 0\end{array}\hfill \end{array}$$

Dividing by $(x+3)$ gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as

$$(x+3)(3x^2+1)$$

We can then set the quadratic equal to 0 and solve to find the other zeros of the function.

$$\begin{array}{l}3x^2+1=0\hfill \\ x^2=-\frac{1}{3}\hfill \\ x=\pm \sqrt{-\frac{1}{3}}=\pm \frac{i\sqrt{3}}{3}\hfill \end{array}$$

The zeros of $f(x)$ are –3 and $\pm \frac{i\sqrt{3}}{3}.$

Analysis

Look at the graph of the function $f$ in Figure 2. Notice that, at $x=-3,$ the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero $x=–3.$ Also note the presence of the two turning points. This means that, since there is a 3^rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of $x=-3$ is 1 and there are two complex solutions, which is what we found, or the multiplicity at $x=-3$ is three. Either way, our result is correct.

Using the Linear Factorization Theorem to Find Polynomials with Given Zeros

A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree $n$ will have $n$ zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into $n$ factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form $(x-c),$ where $c$ is a complex number.

Let $f$ be a polynomial function with real coefficients, and suppose $a+bi\text{, }b\ne 0,$ is a zero of $f(x).$ Then, by the Factor Theorem, $x-(a+bi)$ is a factor of $f(x).$ For $f$ to have real coefficients, $x-(a-bi)$ must also be a factor of $f(x).$ This is true because any factor other than $x-(a-bi),$ when multiplied by $x-(a+bi),$ will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function $f$ with real coefficients has a complex zero $a+bi,$ then the complex conjugate $a-bi$ must also be a zero of $f(x).$ This is called the Complex Conjugate Theorem.

Using Descartes’ Rule of Signs

There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in $f(x)$ and the number of positive real zeros. For example, the polynomial function below has one sign change.

This tells us that the function must have 1 positive real zero.

There is a similar relationship between the number of sign changes in $f(-x)$ and the number of negative real zeros.

In this case, $f(\mathrm{-x})$ has 3 sign changes. This tells us that $f(x)$ could have 3 or 1 negative real zeros.

Example 8

Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for $f(x)=-x^4-3x^3+6x^2-4x-12.$

Solution

Begin by determining the number of sign changes.

There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine $f(-x)$ to determine the number of negative real roots.

$$\begin{array}{l}f(-x)=-{(-x)}^4-3{(-x)}^3+6{(-x)}^2-4(-x)-12\hfill \\ f(-x)=-x^4+3x^3+6x^2+4x-12\hfill \end{array}$$

Again, there are two sign changes, so there are either 2 or 0 negative real roots.

There are four possibilities, as we can see in Table 1.

Table 1 ..

Positive Real Zeros	Negative Real Zeros	Complex Zeros	Total Zeros
2	2	0	4
2	0	2	4
0	2	2	4
0	0	4	4

Analysis

We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure 5. We can see from the graph that the function has 0 positive real roots and 2 negative real roots.

Solving Real-World Applications

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.