4.7: Zeros of Polynomial Functions

Skills to Develop

• 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.

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 \(d(x)\) equals \(x−k\), 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

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

In other words, given the input \(x=k\), the output \(f(k)\) 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 \(x−2\).

\[ 2 \begin{array}{|ccccc} \; 6 & −1 & −15 & 2 & −7 \\ \text{} & 12 & 22 & 14 & 32 \\ \hline \end{array} \\ \begin{array}{ccccc} 6 & 11 & \; 7 & \;\;16 & \;\; 25 \end{array} \nonumber\]

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

Analysis

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

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

\(\PageIndex{1}\)

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

Answer

\(f(−3)=−412\)

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. Look again at the Division Algorithm.

\[f(x)=(x−k)q(x)+r \nonumber\]

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

Written in this form, we can see that \(x−k\) is a factor of \(f(x)\). We can conclude that if \(k\) is a zero of \(f(x)\), then \(x−k\) is a factor of \(f(x)\).

On the other hand, if we know that \(x−k\) is a factor of \(f(x)\), then we can see that the remainder of the Division Algorithm \(f(x)=(x−k)q(x)+r\) is \(0\). Thus, \(k\) 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, it is easy to determine the zeros of the polynomial.

THE FACTOR THEOREM

For a polynomial function \(f(x)\), \(x=k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\).

: Given a factor \(k\) 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 using integer coefficients.
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.

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

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:

\[(x+2)(x^2−8x+15) \nonumber\]

We can factor the quadratic factor to write the polynomial as

\[(x+2)(x−3)(x−5) \nonumber\]

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

\(\PageIndex{2}\)

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.

Answer

The zeros are 2, –2, and –4.

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 (or root) 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 ratios of the factors of the constant term and factors of the leading coefficient of the polynomial.  Note that there may be irrational roots as well.

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, not including 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.

THE RATIONAL ZERO THEOREM

The Rational Zero Theorem states that, if the polynomial \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) 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.

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

1. Determine all factors of the constant term and all factors of the leading coefficient.
2. Determine all possible values of \(\frac{p}{q}\), 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.
3. Determine which possible zeros are actual zeros by evaluating each case of \(f\left(\frac{p}{q}\right)\).

Example \(\PageIndex{3}\): Listing All Possible Rational Zeros

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, divided by the factors of the leading coefficient, 2.

The constant term is –4; the factors of –4 are \(p=±1,±2,±4\).

The leading coefficient is 2; the factors of 2 are \(q=±1,±2\).

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.

\[\dfrac{p}{q}=±\dfrac{1}{1},±\dfrac{1}{2} \qquad \frac{p}{q}=±\dfrac{2}{1},±\dfrac{2}{2} \qquad \dfrac{p}{q}=±\dfrac{4}{1},±\dfrac{4}{2} \nonumber\]

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

\[\dfrac{p}{q} = \dfrac{\text{Factors of the last}}{\text{Factors of the first}}=±1,±2,±4,±\dfrac{1}{2} \nonumber\]

Example \(\PageIndex{4}\): Using the Rational Zero Theorem to Find Rational Zeros

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 \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of 1 and \(q\) is a factor of 2.

\[\dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \nonumber\]

\[=\dfrac{factor\space of\space 1}{factor\space of\space 2} \nonumber\]

The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for \(\frac{p}{q}\) are ±1 and \(±\frac{1}{2}\). 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 \(f(x)\).  Tip: check the value 1 first, because it is so easy: just add up the coefficients.

\[f(−1)=2{(−1)}^3+{(−1)}^2−4(−1)+1=4 \nonumber\]

\[f(1)=1{(1)}^3+{(1)}^2−4(1)+1=0 \nonumber\]

\[f(−\dfrac{1}{2})=2{(−\dfrac{1}{2})}^3+{(−\dfrac{1}{2})}^2−4(−\dfrac{1}{2})+1=3 \nonumber\]

\[f(\dfrac{1}{2})=2{(\dfrac{1}{2})}^3+{(\dfrac{1}{2})}^2−4(\dfrac{1}{2})+1=−\dfrac{1}{2} \nonumber\]

Of those, \(−1\),\(−\dfrac{1}{2}\), and \(\dfrac{1}{2}\) are not zeros of \(f(x)\). 1 is the only rational zero of \(f(x)\).  Keep in mind that there may still be more real roots, that are irrational.  Can you determine if there are?  If so, write your work up clearly and hand it in for Extra Credit.

\(\PageIndex{3}\)

Use the Rational Zero Theorem to find the rational zeros of \(f(x)=x^3−5x^2+2x+1\).

Answer

There are no rational zeros.  However, there may be irrational zeros; in fact, there must be.  If you can clearly explain why, write up your explanation and hand it in for Extra Credit.

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 try to determine all of the zeros of a polynomial function.

: Given a polynomial function \(f\), use synthetic division to find its zeros

1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. 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 and try another.
3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
4. Find the zeros of the quadratic function. Three possible methods for solving quadratics are factoring, completing the square, or using the quadratic formula.

Example \(\PageIndex{5}\): 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 \(\dfrac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of –1 and \(q\) is a factor of 4.

\[\dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \nonumber\]

\[=\dfrac{factor\space of\space -1}{factor\space of\space 4} \nonumber\]

The factors of –1 are ±1 and the factors of 4 are ±1, ±2, and ±4. The possible values for \(\dfrac{p}{q}\) are \(±1\), \(±\frac{1}{2}\), and \(±\frac{1}{4}\). 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.  It is easy to determine that 1 is a zero, since adding the coefficients of each term gives \(4+(-3)+(-1)=0\). However, we will still use synthetic division in order to write the polynomial in factored form.

Dividing by \((x−1)\) gives a remainder of 0, as expected. The polynomial can be written as

\[(x−1)(4x^2+4x+1) = (x-1)((2x)^2 +4x +1^2) \nonumber\]

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

\[(x−1){(2x+1)}^2 \nonumber\]

We know that 1 is a zero. To find the other zero, we can set the second factor equal to 0.

\[(2x+1)^2=0 \nonumber\]

\[x=−\dfrac{1}{2} \nonumber\]

The zeros of the function are 1 (with multiplicity 1) and \(−\frac{1}{2}\) (with multiplicity 2).

Analysis

Look at the graph of the function \(f\) in Figure \(\PageIndex{1}\). Notice, at \(x =−0.5\), the graph bounces off the \(x\)-axis, indicating the even multiplicity for the zero −0.5. At \(x=1\), the graph crosses the \(x\)-axis, indicating the odd multiplicity for the zero \(x=1\).

Figure \(\PageIndex{1}\).

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.  (Keep in mind that real numbers are complex numbers of the form \(0+bi\).) 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 \(f(x)\).

THE FUNDAMENTAL THEOREM OF ALGEBRA

The Fundamental Theorem of Algebra states that, 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

\[f(x)=a(x−c_1)(x−c_2)...(x−c_n) \nonumber\]

where \(c_1,c_2\),...,\(c_n\) are complex numbers. Therefore, \(f(x)\) has \(n\) complex roots if we count all multiplicities.

Does every polynomial have at least one imaginary zero?

No. Real numbers are a subset of complex numbers, but not the other way around. A complex number does not necessarily have an imaginary part. Real numbers are also complex numbers.

Example \(\PageIndex{6}\): 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.

\[\dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \nonumber\]

\[=\dfrac{factor\space of\space 3}{factor\space of\space 3} \nonumber\]

The factors of 3 are ±1 and ±3. The possible values for \(\dfrac{p}{q}\), and therefore the possible rational zeros for the function, are ±3, ±1, and \(±\dfrac{1}{3}\). 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 (checking quickly that 1 is not a zero).

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

\[(x+3)(3x^2+1).\]

(Note: this polynomial could also have been factored using the technique of grouping.)

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

\[3x^2+1=0 \nonumber\]

\[x^2=−\dfrac{1}{3} \nonumber\]

\[x=±\sqrt{-\dfrac{1}{3}}=±\dfrac{i\sqrt{3}}{3} \nonumber\]

The zeros of \(f(x)\) are \(–3\) and \(±\dfrac{i\sqrt{3}}{3}\).

Analysis

Suppose we do not have the formula for the function, but we have a graph of the function. Look at the graph of the function \(f(x)=3x^3+9x^2+x+3\) in Figure \(\PageIndex{2}\). Notice that, at \(x =−3\), the graph crosses the \(x\)-axis, indicating an odd multiplicity 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. 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.  Without only a graph to look at, we are not able to determine which is the correct choice, so either answer is acceptable.

Figure \(\PageIndex{2}\).

\(\PageIndex{4}\)

Find the zeros of \(f(x)=2x^3+5x^2−11x+4\).

Answer

The zeros are \(–4\), \(\frac{1}{2}\), and \(1\).

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 states 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 \(a +bi\), \(b≠0\), 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.

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.

: 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

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

Example \(\PageIndex{7}\): Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros

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.

\[\begin{align*} f(x) & =a(x+3)(x−2)(x−i)(x+i) \\ f(x) & =a(x^2+x−6)(x^2+1) \\ f(x) & =a(x^4+x^3−5x^2+x−6) \end{align*} \]

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

\[\begin{align*} 100=a({(−2)}^4+{(−2)}^3−5{(−2)}^2+(−2)−6) \\ 100=a(−20) \\ −5=a \end{align*} \]

So the polynomial function is

\[f(x)=−5(x^4+x^3−5x^2+x−6), \nonumber\]

which can be written in general form as

\[f(x)=−5x^4−5x^3+25x^2−5x+30. \nonumber\]

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 \(i\) .

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.

\(\PageIndex{5}\)

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

Answer

\(f(x)=−\frac{1}{2}x^3+\frac{5}{2}x^2−2x+10\)

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 general form, 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, \(f(−x)\) has 3 sign changes. This tells us (see below) that \(f(x)\) could have 3 or 1 negative real zeros.

DESCARTES’ RULE OF SIGNS

According to Descartes’ Rule of Signs, if we let \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) 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 \(\PageIndex{8}\): 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.

\[ \begin{align*} f(−x) & =−{(−x)}^4−3{(−x)}^3+6{(−x)}^2−4(−x)−12 \\ f(−x) & =−x^4+3x^3+6x^2+4x−12 \end{align*} \]

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 \(\PageIndex{1}\).

Table \(\PageIndex{1}\)

Positive Real Zeros	Negative Real Zeros	Zeros with an Imaginary Part	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 \(\PageIndex{3}\). We can see from the graph that the function has 0 positive real roots and 2 negative real roots.

Figure \(\PageIndex{3}\).

\(\PageIndex{6}\)

Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for \(f(x)=2x^4−10x^3+11x^2−15x+12\). Use a graph to verify the numbers of positive and negative real zeros for the function.

Answer

There must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.

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 \(\PageIndex{9}\) A POLYNOMIAL APPLICATION

 

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 \(V=lwh\). 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 \(h=\dfrac{1}{3}w\). Let’s write the volume of the cake in terms of width of the cake.

\[V=(w+4)(w)(\dfrac{1}{3}w) \nonumber\]

\[V=\dfrac{1}{3}w^3+\dfrac{4}{3}w^2 \nonumber\]

Substitute the given volume into this equation.

\(351=\frac{1}{3}w^3+\frac{4}{3}w^2\) Substitute 351 for V.

\(1053=w^3+4w^2\) Multiply both sides by 3.

\(0=w^3+7w^2−1053\) Subtract 1053 from both sides.

 

Descartes' Rule of Signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are \(\pm 1,±3,±9,±13,±27,±39,±81,±117,±351,\) and \(±1053\). 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. A 1-inch wide cake is not likely, and by adding the coefficients, we can see that 1 is not a solution.

We will use synthetic division to check \(x=3\).

Since 3 is not a solution either, we will test \(x=9\).

Synthetic division gives a remainder of 0, so 9 is the one positive 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.

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

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

 

\(\PageIndex{7}\)

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, 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. What should the dimensions of the container be?

Answer

3 meters by 4 meters by 7 meters

Media

Access these online resources for additional instruction and practice with zeros of polynomial functions.

• Real Zeros, Factors, and Graphs of Polynomial Functions
• Complex Factorization Theorem
• Find the Zeros of a Polynomial Function
• Find the Zeros of a Polynomial Function 2
• Find the Zeros of a Polynomial Function 3

Key Concepts

• To find \(f(k)\), determine the remainder of the polynomial \(f(x)\) when it is divided by \(x−k\). This is known as the Remainder Theorem. See Example \(\PageIndex{1}\).
• According to the Factor Theorem, \(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\). See Example\(\PageIndex{2}\).
• According to the Rational Zero Theorem, 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. See Example \(\PageIndex{3}\) and Example \(\PageIndex{4}\).
• 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. See Example \(\PageIndex{5}\).
• According to the Fundamental Theorem, every polynomial function with degree greater than 0 has at least one complex zero (keep in mind that real numbers are complex numbers). See Example \(\PageIndex{6}\)., as well as Examples \(\PageIndex{2}\), \(\PageIndex{4}\), and \(\PageIndex{5}\).
• 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. See Example \(\PageIndex{7}\).
• 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. See Example \(\PageIndex{8}\).
• Polynomial equations model many real-world scenarios. Synthetic division can often be used to solve the equation. See Example \(\PageIndex{9}\).

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 \(f(x)\) 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 \(\frac{p}{q}\)​​​​​​​ 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 \(f(k)\)

Contributor

• Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.