Zeros of Polynomial Functions

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.

Evaluate 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

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.

 Use 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 [latex]f\left(k\right)=0[/latex] and [latex]f\left(x\right)=\left(x-k\right)q\left(x\right)+0[/latex] or [latex]f\left(x\right)=\left(x-k\right)q\left(x\right)[/latex].

Notice, written in this form, x – k is a factor of [latex]f\left(x\right)[/latex]. We can conclude if k is a zero of [latex]f\left(x\right)[/latex], then [latex]x-k[/latex] is a factor of [latex]f\left(x\right)[/latex].

Similarly, if [latex]x-k[/latex] is a factor of [latex]f\left(x\right)[/latex], then the remainder of the Division Algorithm [latex]f\left(x\right)=\left(x-k\right)q\left(x\right)+r[/latex] 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.

Example 2: Using the Factor Theorem to Solve a Polynomial Equation

Show that [latex]\left(x+2\right)[/latex] is a factor of [latex]{x}^{3}-6{x}^{2}-x+30[/latex]. Find the remaining factors. Use the factors to determine the zeros of the polynomial.

Solutions

We can use synthetic division to show that [latex]\left(x+2\right)[/latex] is a factor of the polynomial.

The remainder is zero, so [latex]\left(x+2\right)[/latex] 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:

[latex]\left(x+2\right)\left({x}^{2}-8x+15\right)[/latex]

We can factor the quadratic factor to write the polynomial as

[latex]\left(x+2\right)\left(x - 3\right)\left(x - 5\right)[/latex]

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

 Use 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, [latex]x=\frac{2}{5}[/latex] and [latex]x=\frac{3}{4}[/latex].

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.

 Find zeros of a polynomial function

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 [latex]f\left(x\right)=4{x}^{3}-3x - 1[/latex].

Solution

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

[latex]\begin{cases}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ \text{ }=\frac{\text{factor of -1}}{\text{factor of 4}}\hfill \end{cases}[/latex]

The factors of –1 are [latex]\pm 1[/latex] and the factors of 4 are [latex]\pm 1,\pm 2[/latex], and [latex]\pm 4[/latex]. The possible values for [latex]\frac{p}{q}[/latex] are [latex]\pm 1,\pm \frac{1}{2}[/latex], and [latex]\pm \frac{1}{4}[/latex]. 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 [latex]\left(x - 1\right)[/latex] gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as

[latex]\left(x - 1\right)\left(4{x}^{2}+4x+1\right)[/latex].

The quadratic is a perfect square. [latex]f\left(x\right)[/latex] can be written as

[latex]\left(x - 1\right){\left(2x+1\right)}^{2}[/latex].

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.

[latex]\begin{cases}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{cases}[/latex]

The zeros of the function are 1 and [latex]-\frac{1}{2}[/latex] with multiplicity 2.

Analysis of the Solution

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

Figure 1

 Use 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 [latex]f\left(x\right)=0[/latex]. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it [latex]{c}_{1}[/latex]. By the Factor Theorem, we can write [latex]f\left(x\right)[/latex] as a product of [latex]x-{c}_{\text{1}}[/latex] and a polynomial quotient. Since [latex]x-{c}_{\text{1}}[/latex] 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 [latex]{c}_{\text{2}}[/latex]. So we can write the polynomial quotient as a product of [latex]x-{c}_{\text{2}}[/latex] 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 [latex]f\left(x\right)[/latex].

Example 6: Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of [latex]f\left(x\right)=3{x}^{3}+9{x}^{2}+x+3[/latex].

Solution

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

[latex]\begin{cases}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ \text{ }=\frac{\text{factor of 3}}{\text{factor of 3}}\hfill \end{cases}[/latex]

The factors of 3 are [latex]\pm 1[/latex] and [latex]\pm 3[/latex]. The possible values for [latex]\frac{p}{q}[/latex], and therefore the possible rational zeros for the function, are [latex]\pm 3,\text{\pm 1, and }\pm \frac{1}{3}[/latex]. 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.

Dividing by [latex]\left(x+3\right)[/latex] gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as

[latex]\left(x+3\right)\left(3{x}^{2}+1\right)[/latex]

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

[latex]\begin{cases}3{x}^{2}+1=0\hfill \\ \text{ }{x}^{2}=-\frac{1}{3}\hfill \\ \text{ }x=\pm \sqrt{-\frac{1}{3}}=\pm \frac{i\sqrt{3}}{3}\hfill \end{cases}[/latex]

The zeros of [latex]f\left(x\right)[/latex] are –3 and [latex]\pm \frac{i\sqrt{3}}{3}[/latex].

Analysis of the Solution

Look at the graph of the function f. Notice that, at [latex]x=-3[/latex], the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[/latex]. 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 [latex]x=-3[/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[/latex] is three. Either way, our result is correct.

Figure 2

 Use 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 [latex]a+bi\text{, }b\ne 0[/latex], is a zero of [latex]f\left(x\right)[/latex]. Then, by the Factor Theorem, [latex]x-\left(a+bi\right)[/latex] is a factor of [latex]f\left(x\right)[/latex]. For f to have real coefficients, [latex]x-\left(a-bi\right)[/latex] must also be a factor of [latex]f\left(x\right)[/latex]. This is true because any factor other than [latex]x-\left(a-bi\right)[/latex], when multiplied by [latex]x-\left(a+bi\right)[/latex], 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 [latex]a+bi[/latex], then the complex conjugate [latex]a-bi[/latex] must also be a zero of [latex]f\left(x\right)[/latex]. This is called the Complex Conjugate Theorem.

 Use 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 [latex]f\left(x\right)[/latex] and the number of positive real zeros. For example, the polynomial function below has one sign change.

There is a similar relationship between the number of sign changes in [latex]f\left(-x\right)[/latex] and the number of negative real zeros.

Example 7: Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for [latex]f\left(x\right)=-{x}^{4}-3{x}^{3}+6{x}^{2}-4x - 12[/latex].

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 [latex]f\left(-x\right)[/latex] to determine the number of negative real roots.

[latex]\begin{cases}f\left(-x\right)=-{\left(-x\right)}^{4}-3{\left(-x\right)}^{3}+6{\left(-x\right)}^{2}-4\left(-x\right)-12\hfill \\ f\left(-x\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12\hfill \end{cases}[/latex]

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

There are four possibilities, as we can see below.

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 of the Solution

We can confirm the numbers of positive and negative real roots by examining a graph of the function. We can see from the graph in Figure 3 that the function has 0 positive real roots and 2 negative real roots.

Figure 3

 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?

Solution

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.

[latex]\begin{cases}V=\left(w+4\right)\left(w\right)\left(\frac{1}{3}w\right)\\ V=\frac{1}{3}{w}^{3}+\frac{4}{3}{w}^{2}\end{cases}[/latex]

Substitute the given volume into this equation.

[latex]\begin{cases}\text{ }351=\frac{1}{3}{w}^{3}+\frac{4}{3}{w}^{2}\hfill & \text{Substitute 351 for }V.\hfill \\ 1053={w}^{3}+4{w}^{2}\hfill & \text{Multiply both sides by 3}.\hfill \\ \text{ }0={w}^{3}+4{w}^{2}-1053 \hfill & \text{Subtract 1053 from both sides}.\hfill \end{cases}[/latex]

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.

[latex]l=w+4=9+4=13\text{ and }h=\frac{1}{3}w=\frac{1}{3}\left(9\right)=3[/latex]

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

Key Concepts

• To find [latex]f\left(k\right)[/latex], determine the remainder of the polynomial [latex]f\left(x\right)[/latex] when it is divided by [latex]x-k[/latex].
• k is a zero of [latex]f\left(x\right)[/latex] if and only if [latex]\left(x-k\right)[/latex] is a factor of [latex]f\left(x\right)[/latex].
• Each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient.
• When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
• Synthetic division can be used to find the zeros of a polynomial function.
• According to the Fundamental Theorem, every polynomial function has at least one complex zero.
• Every polynomial function with degree greater than 0 has at least one complex zero.
• Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form [latex]\left(x-c\right)[/latex], where c is a complex number.
• The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer.
• The number of negative real zeros of a polynomial function is either the number of sign changes of [latex]f\left(-x\right)[/latex] or less than the number of sign changes by an even integer.
• Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division.

Glossary

Descartes’ Rule of Signs

a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of [latex]f\left(x\right)[/latex] and [latex]f\left(-x\right)[/latex]

Factor Theorem

k is a zero of polynomial function [latex]f\left(x\right)[/latex] if and only if [latex]\left(x-k\right)[/latex] is a factor of [latex]f\left(x\right)[/latex]

Fundamental Theorem of Algebra

a polynomial function with degree greater than 0 has at least one complex zero

Linear Factorization Theorem

allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\left(x-c\right)[/latex], where c is a complex number

Rational Zero Theorem

the possible rational zeros of a polynomial function have the form [latex]\frac{p}{q}[/latex] where p is a factor of the constant term and q is a factor of the leading coefficient.

Remainder Theorem

if a polynomial [latex]f\left(x\right)[/latex] is divided by [latex]x-k[/latex], then the remainder is equal to the value [latex]f\left(k\right)[/latex]

Section Exercises

Licenses & Attributions