Zeros of Polynomial Functions

By the end of this section, you will be able to:

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, multi-tiered cakes for display and cutting at Quinceañera and wedding celebrations, as well as sheet cakes to serve most of the guests. The bakery wants the volume of a small sheet 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)

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{array}{lrcl} & f(k) &=& (k-k)q(k)+r \\[4pt] & &=& 0\cdot q(k)+r \\[4pt] & &=& r \end{array}

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.

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

Example. 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 $x-2$ .

\begin{array}{r|rrrrr} 2 & 6 & -1 & -15 & 2 & -7 \\ & & 12 & 22 & 14 & 32 \\ \hline & 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{array}{lrcl} & f(x) &=& 6x^4-x^3-15x^2+2x-7 \\[4pt] & f(2) &=& 6(2)^4-(2)^3-15(2)^2+2(2)-7 \\[4pt] & &=& 25 \end{array}

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

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

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.

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

Example. 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.

\begin{array}{r|rrrr} -2 & 1 & -6 & -1 & 30 \\ & & -2 & 16 & -30 \\ \hline & 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.

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. Enter all of them, separated by commas.

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, $x=\tfrac{2}{5}$ and $x=\tfrac{3}{4}$ . 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}{lrcl} \text{Set each factor equal to }0. & x-\tfrac{2}{5} &=& 0\ \text{or}\ x-\tfrac{3}{4}=0 \\[4pt] \text{Multiply both sides by the denominator to clear fractions.} & 5x-2 &=& 0\ \text{or}\ 4x-3=0 \\[4pt] \text{Create the quadratic function, multiplying the factors.} & f(x) &=& (5x-2)(4x-3) \\[4pt] \text{Expand the polynomial.} & f(x) &=& 20x^2-23x+6 \\[4pt] & f(x) &=& (5\cdot4)x^2-23x+(2\cdot3) \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 $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ has integer coefficients and $a_n\ne0$ , then every rational zero of $f(x)$ has the form $\tfrac{p}{q}$ 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.

Determine all factors of the constant term and all factors of the leading coefficient.
Determine all possible values of $\tfrac{p}{q}$ , where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. Be sure to include both positive and negative candidates.
Determine which possible zeros are actual zeros by evaluating each case of $f\left(\tfrac{p}{q}\right)$ .

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

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 $p=\pm1,\pm2,\pm4$ .

The leading coefficient is 2; the factors of 2 are $q=\pm1,\pm2$ .

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

\begin{array}{lrcl} & \tfrac{p}{q} &=& \pm\tfrac{1}{1},\ \pm\tfrac{1}{2} \\[4pt] & \tfrac{p}{q} &=& \pm\tfrac{2}{1},\ \pm\tfrac{2}{2} \\[4pt] & \tfrac{p}{q} &=& \pm\tfrac{4}{1},\ \pm\tfrac{4}{2} \end{array}

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

\tfrac{p}{q}=\tfrac{\text{Factors of the last}}{\text{Factors of the first}}=\pm1,\pm2,\pm4,\pm\tfrac{1}{2}

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

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

\begin{array}{lrcl} & \tfrac{p}{q} &=& \tfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[4pt] & &=& \tfrac{\text{factor of 1}}{\text{factor of 2}} \end{array}

The factors of 1 are $\pm1$ and the factors of 2 are $\pm1$ and $\pm2$ . The possible values for $\tfrac{p}{q}$ are $\pm1$ and $\pm\tfrac{1}{2}$ . 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{array}{lrcl} & f(-1) &=& 2(-1)^3+(-1)^2-4(-1)+1=4 \\[4pt] & f(1) &=& 2(1)^3+(1)^2-4(1)+1=0 \\[4pt] & f\left(-\tfrac{1}{2}\right) &=& 2\left(-\tfrac{1}{2}\right)^3+\left(-\tfrac{1}{2}\right)^2-4\left(-\tfrac{1}{2}\right)+1=3 \\[4pt] & f\left(\tfrac{1}{2}\right) &=& 2\left(\tfrac{1}{2}\right)^3+\left(\tfrac{1}{2}\right)^2-4\left(\tfrac{1}{2}\right)+1=-\tfrac{1}{2} \end{array}

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

Use the Rational Zero Theorem to determine which of the following is true about the rational zeros of $f(x)=x^3-5x^2+2x+1$ .

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.

How to: given a polynomial function $f$ , use synthetic division to find its zeros.

Use the Rational Zero Theorem to list all possible rational zeros of the function.
Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.

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

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

\begin{array}{lrcl} & \tfrac{p}{q} &=& \tfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[4pt] & &=& \tfrac{\text{factor of }-1}{\text{factor of 4}} \end{array}

The factors of $-1$ are $\pm1$ and the factors of 4 are $\pm1,\pm2,$ and $\pm4$ . The possible values for $\tfrac{p}{q}$ are $\pm1$ , $\pm\tfrac{1}{2}$ , and $\pm\tfrac{1}{4}$ . 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}{r|rrrr} 1 & 4 & 0 & -3 & -1 \\ & & 4 & 4 & 1 \\ \hline & 4 & 4 & 1 & 0 \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}{lrcl} & 2x+1 &=& 0 \\[4pt] & x &=& -\tfrac{1}{2} \end{array}

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

Analysis. Look at the graph of the function $f$ below. 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: The graph of f of x equals 4x cubed minus 3x minus 1, on a grid with x from negative 1.8 to 1.8 and y from negative 2.3 to 1.5. The curve rises from the lower left, touches the x-axis at x equals negative 0.5 without crossing it, labeled Bounce, dips to a local minimum near x equals 0.5, then rises again to cross the x-axis at x equals 1, labeled Cross, and continues upward.

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 $x-c_1$ and a polynomial quotient. Since $x-c_1$ 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 $c_2$ . So we can write the polynomial quotient as a product of $x-c_2$ 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 $f(x)$ .

Fundamental Theorem of Algebra. If $f(x)$ is a polynomial of degree $n>0$ , then $f(x)$ has at least one complex zero.

We can use this theorem to argue that, if $f(x)$ is a polynomial of degree $n>0$ , and $a$ is a non-zero real number, then $f(x)$ has exactly $n$ linear factors

f(x)=a(x-c_1)(x-c_2)\ldots(x-c_n)

where $c_1,c_2,\ldots,c_n$ are complex numbers. Therefore, $f(x)$ has $n$ roots if we allow for multiplicities.

Q&A. Does every polynomial have at least one imaginary zero?

No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

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

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

\begin{array}{lrcl} & \tfrac{p}{q} &=& \tfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[4pt] & &=& \tfrac{\text{factor of 3}}{\text{factor of 3}} \end{array}

The factors of 3 are $\pm1$ and $\pm3$ . The possible values for $\tfrac{p}{q}$ , and therefore the possible rational zeros for the function, are $\pm3$ , $\pm1$ , and $\pm\tfrac{1}{3}$ . 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}{r|rrrr} -3 & 3 & 9 & 1 & 3 \\ & & -9 & 0 & -3 \\ \hline & 3 & 0 & 1 & 0 \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}{lrcl} & 3x^2+1 &=& 0 \\[4pt] & x^2 &=& -\tfrac{1}{3} \\[4pt] & x &=& \pm\sqrt{-\tfrac{1}{3}}=\pm\tfrac{i\sqrt{3}}{3} \end{array}

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

Analysis. Look at the graph of the function $f$ below. 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 3rd 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.

Figure: The graph of f of x equals 3x cubed plus 9x squared plus x plus 3, on a grid with x from negative 4.3 to 2.3 and y from negative 3.5 to 17. The curve rises from the lower left, crosses the x-axis at x equals negative 3, labeled Cross, rises to a local maximum near x equals negative 1.9, dips to a local minimum near x equals 0 without recrossing the axis, then rises steeply to the right.

Find the zeros of $f(x)=2x^3+5x^2-11x+4$ . Enter all of them, separated by commas.

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 $a+bi$ , $b\ne0$ , 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.

Complex Conjugate Theorem. According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form $(x-c)$ , where $c$ is a complex number.

If the polynomial function $f$ has real coefficients and a complex zero in the form $a+bi$ , then the complex conjugate of the zero, $a-bi$ , is also a zero.

How to: given the zeros of a polynomial function $f$ and a point $(c,f(c))$ on the graph of $f$ , use the Linear Factorization Theorem to find the polynomial function.

Use the zeros to construct the linear factors of the polynomial.
Multiply the linear factors to expand the polynomial.
Substitute $(c,f(c))$ into the function to determine the leading coefficient.
Simplify.

Example. Find a fourth degree polynomial with real coefficients that has zeros of $-3$ , 2, $i$ , such that $f(-2)=100$ .

Solution. Because $x=i$ is a zero, by the Complex Conjugate Theorem $x=-i$ is also a zero. The polynomial must have factors of $(x+3)$ , $(x-2)$ , $(x-i)$ , and $(x+i)$ . Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.

\begin{array}{lrcl} & f(x) &=& a(x+3)(x-2)(x-i)(x+i) \\[4pt] & f(x) &=& a(x^2+x-6)(x^2+1) \\[4pt] & f(x) &=& a(x^4+x^3-5x^2+x-6) \end{array}

We need to find $a$ to ensure $f(-2)=100$ . Substitute $x=-2$ and $f(-2)=100$ into $f(x)$ .

\begin{array}{lrcl} & 100 &=& a\bigl((-2)^4+(-2)^3-5(-2)^2+(-2)-6\bigr) \\[4pt] & 100 &=& a(-20) \\[4pt] & -5 &=& a \end{array}

So the polynomial function is

f(x)=-5(x^4+x^3-5x^2+x-6)

f(x)=-5x^4-5x^3+25x^2-5x+30

Analysis. We found that both $i$ and $-i$ were zeros, but only one of these zeros needed to be given. If $i$ is a zero of a polynomial with real coefficients, then $-i$ must also be a zero of the polynomial because $-i$ is the complex conjugate of $i$ .

Q&A. If $2+3i$ were given as a zero of a polynomial with real coefficients, would $2-3i$ also need to be a zero?

Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.

Find a third-degree polynomial function with real coefficients that has zeros of 5 and $-2i$ such that $f(1)=10$ .

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.

f(x)=x^4+x^3+x^2+x-1

The sign changes once, from $x$ to $-1$ . 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.

\begin{array}{lrcl} & f(-x) &=& (-x)^4+(-x)^3+(-x)^2+(-x)-1 \\[4pt] & f(-x) &=& x^4-x^3+x^2-x-1 \end{array}

The sign changes three times: from $x^4$ to $-x^3$ , from $-x^3$ to $x^2$ , and from $x^2$ to $-x$ . In this case, $f(-x)$ has 3 sign changes. This tells us that $f(x)$ could have 3 or 1 negative real zeros.

Descartes’ Rule of Signs. According to Descartes’ Rule of Signs, if we let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ be a polynomial function with real coefficients:

The number of positive real zeros is either equal to the number of sign changes of $f(x)$ or is less than the number of sign changes by an even integer.
The number of negative real zeros is either equal to the number of sign changes of $f(-x)$ or is less than the number of sign changes by an even integer.

Example. 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.

f(x)=-x^4-3x^3+6x^2-4x-12

The sign changes twice: from $-3x^3$ to $6x^2$ , and from $6x^2$ to $-4x$ . 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}{lrcl} & f(-x) &=& -(-x)^4-3(-x)^3+6(-x)^2-4(-x)-12 \\[4pt] & f(-x) &=& -x^4+3x^3+6x^2+4x-12 \end{array}

The sign changes twice: from $-x^4$ to $3x^3$ , and from $4x$ to $-12$ . 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 the table 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. We can confirm the numbers of positive and negative real roots by examining a graph of the function below. We can see from the graph that the function has 0 positive real roots and 2 negative real roots.

Figure: The graph of f of x equals negative x to the fourth minus 3x cubed plus 6x squared minus 4x minus 12, on a grid with x from negative 5.3 to 5.3 and y from negative 33 to 63. The curve rises from the lower left, crosses the x-axis near x equals negative 4.42, rises to a local maximum, falls and crosses the x-axis again at x equals negative 1, then dips to a local minimum and continues downward to the lower right without crossing the axis again, confirming zero positive real zeros and two negative real zeros.

Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros of $f(x)=2x^4-10x^3+11x^2-15x+12$ .

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.

Example. A new bakery offers decorated, multi-tiered cakes for display and cutting at Quinceañera and wedding celebrations, as well as sheet cakes to serve most of the guests. The bakery wants the volume of a small sheet 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 $V=lwh$ . We were given that the length must be four inches longer than the width, so we can express the length of the cake as $l=w+4$ . 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 $h=\tfrac{1}{3}w$ . Let’s write the volume of the cake in terms of width of the cake.

\begin{array}{lrcl} & V &=& (w+4)(w)\left(\tfrac{1}{3}w\right) \\[4pt] & V &=& \tfrac{1}{3}w^3+\tfrac{4}{3}w^2 \end{array}

Substitute the given volume into this equation.

\begin{array}{lrcl} \text{Substitute 351 for }V. & 351 &=& \tfrac{1}{3}w^3+\tfrac{4}{3}w^2 \\[4pt] \text{Multiply both sides by 3}. & 1053 &=& w^3+4w^2 \\[4pt] \text{Subtract 1053 from both sides}. & 0 &=& w^3+4w^2-1053 \end{array}

Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are $\pm1,\pm3,\pm9,\pm13,\pm27,\pm39,\pm81,\pm117,\pm351,$ and $\pm1053$ . 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 $x=1$ .

\begin{array}{r|rrrr} 1 & 1 & 4 & 0 & -1053 \\ & & 1 & 5 & 5 \\ \hline & 1 & 5 & 5 & -1048 \end{array}

Since 1 is not a solution, we will check $x=3$ .

\begin{array}{r|rrrr} 3 & 1 & 4 & 0 & -1053 \\ & & 3 & 21 & 63 \\ \hline & 1 & 7 & 21 & -990 \end{array}

Since 3 is not a solution either, we will test $x=9$ .

\begin{array}{r|rrrr} 9 & 1 & 4 & 0 & -1053 \\ & & 9 & 117 & 1053 \\ \hline & 1 & 13 & 117 & 0 \end{array}

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.

l=w+4=9+4=13\quad\text{and}\quad h=\tfrac{1}{3}w=\tfrac{1}{3}(9)=3

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

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. Enter the width, length, and height, in meters, in that order, separated by commas.

Key concepts

To find $f(k)$ , determine the remainder of the polynomial $f(x)$ when it is divided by $x-k$ .
$k$ is a zero of $f(x)$ if and only if $(x-k)$ is a factor of $f(x)$ .
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 $(x-c)$ , 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 $f(-x)$ 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.

Key terms

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 $f(x)$ and $f(-x)$ . Factor Theorem — $k$ is a zero of polynomial function $f(x)$ if and only if $(x-k)$ is a factor of $f(x)$ . 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 $(x-c)$ , where $c$ is a complex number. Rational Zero Theorem — the possible rational zeros of a polynomial function have the form $\tfrac{p}{q}$ where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. Remainder Theorem — if a polynomial $f(x)$ is divided by $x-k$ , then the remainder is equal to the value $f(k)$ .

Practice

Evaluate a polynomial using the Remainder Theorem

Use the Remainder Theorem to find the remainder when $3x^3-2x^2+x-4$ is divided by $x+3$ .

Use the Remainder Theorem to find the remainder when $-3x^2+6x+24$ is divided by $x-4$ .

Use the Remainder Theorem to find the remainder when $x^4-1$ is divided by $x-4$ .

Use the Factor Theorem to solve a polynomial equation

Use the given factor and the Factor Theorem to find all real zeros of $f(x)=2x^3+x^2-5x+2$ ; $x+2$ . Enter all of them, separated by commas.

Use the given factor and the Factor Theorem to find all real zeros of $2x^3+5x^2-12x-30$ ; $2x+5$ . Enter all of them, separated by commas.

Use the Rational Zero Theorem to find rational zeros

Use the Rational Zero Theorem to list all possible rational zeros of $f(x)=2x^3+3x^2-8x+5$ . Enter all of them, separated by commas.

Use the Rational Zero Theorem to list all possible rational zeros of $f(x)=6x^4-10x^2+13x+1$ . Enter all of them, separated by commas.

Find zeros of a polynomial function

Use the Rational Zero Theorem to find all real zeros of $2x^3+7x^2-10x-24=0$ . Enter all of them, separated by commas.

Find all complex solutions (real and non-real) of $x^3-8x^2+25x-26=0$ . Enter all of them, separated by commas.

Use a graph to find the rational zeros of $f(x)=6x^3-7x^2+1$ . Enter all of them, separated by commas.

Use the Linear Factorization Theorem to find polynomials with given zeros

Construct a polynomial function of least degree possible with real roots $-1$ (multiplicity 2), $1$ (multiplicity 1), and $(2,f(2))=(2,4)$ .

Construct a polynomial function of least degree possible with real roots $-\tfrac12$ , $0$ , $\tfrac12$ , and $(-2,f(-2))=(-2,6)$ .

Use Descartes’ Rule of Signs

According to Descartes’ Rule of Signs, which describes the possible numbers of positive and negative real zeros of $f(x)=x^4-x^2-1$ ?

According to Descartes’ Rule of Signs, which describes the possible numbers of positive and negative real zeros of $f(x)=x^3-2x^2+x-1$ ?

According to Descartes’ Rule of Signs, which describes the possible numbers of positive and negative real zeros of $f(x)=2x^3+37x^2+200x+300$ ?

Solve real-world applications of polynomial equations

A box’s length is twice its width, and its height is 2 inches greater than its width. The volume is 192 cubic inches. Enter the length, width, and height, in inches, in that order, separated by commas.

A box’s length is one inch more than its width, which is one inch more than its height. The volume is $86.625$ cubic inches. Enter the length, width, and height, in inches, in that order, separated by commas.

This section is adapted from Precalculus 2e, Section 3.6: Zeros of Polynomial Functions by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the media links list of external practice resources; recreated the introductory Rational Zero Theorem derivation image (setting the two fractional zeros’ factors equal to 0, clearing denominators, and expanding to $20x^2-23x+6$ ) as a KaTeX step array; recreated the four sign-change annotation images in the Descartes’ Rule of Signs discussion (the introductory example and Example 8’s $f(x)$ and $f(-x)$ ) as displayed equations with the sign-changing term pairs named in prose, since the figure toolkit has no primitive for arc annotations on a typeset equation; recreated the two synthetic-division stage images completing the bakery example’s third and fourth tests ( $x=3$ and $x=9$ ) as KaTeX synthetic-division arrays matching the first stage already shown; recreated the three instructional graphs (the bounce/cross graph of $f(x)=4x^3-3x-1$ , the cross graph of $f(x)=3x^3+9x^2+x+3$ , and the Descartes-confirming graph of $f(x)=-x^4-3x^3+6x^2-4x-12$ ) as accessible SVGs plotted from the exact polynomials coefficients rather than as images; omitted the five answer-key confirmation graphs accompanying the “Graphical” end-of-section exercises drawn into the Descartes’ Rule of Signs Practice group, since each component’s graded answer is the source’s own printed Descartes-analysis text (not a graph reading) and Example 8 already demonstrates the graph-confirmation step in full; converted every “Try It” into an interactive component with instant feedback, with the “no rational zeros” and Descartes’ Rule Try Its built as multiple-choice because their answers are a declarative fact or a described set of possibilities rather than a single value; and adapted 17 selected end-of-section exercises into interactive Practice components, one or more per objective group, presenting each multi-dimension word-problem answer in the length/width/height (or width/length/height) order the source’s own solution used, stated explicitly in the question.

Dividing Polynomials Rational Functions