3.6 Zeros of Polynomial Functions - Precalculus 2e | OpenStax (2024)

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

$$\begin{array}{l}\\ 2\overline{)\begin{array}{lllll}6\hfill & -1\hfill & -15\hfill & 2\hfill & -7\hfill \\ \hfill & 12\hfill & 22\hfill & 14\hfill & 32\hfill \end{array}}\\ \begin{array}{lllll}6\hfill & 11\hfill & 7\hfill & \text{16}\hfill & 25\hfill \end{array}\end{array}$$

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

We can use synthetic division to show that $(x+2)$ is a factor of the polynomial.

$$\begin{array}{l}\hfill \\ \begin{array}{l}-2\hfill \\ \hfill \end{array}\overline{)\begin{array}{rrrr}\hfill 1& \hfill -6& \hfill -1& \hfill 30\\ \hfill & \hfill -2& \hfill 16& \hfill -30\end{array}}\hfill \\ \begin{array}{rrrr}\hfill \text{1}& \hfill -8& \hfill 15& \hfill 0\end{array}\hfill \end{array}$$

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.

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 $p=\mathrm{\pm 1},\mathrm{\pm 2},\mathrm{\pm 4.}$

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

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

$$\begin{array}{ccc}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2}& \frac{p}{q}=\pm \frac{2}{1},\pm \frac{2}{2}& \frac{p}{q}=\pm \frac{4}{1},\pm \frac{4}{2}\end{array}$$

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

$$\frac{p}{q}=\frac{\text{Factorsofthelast}}{\text{Factorsofthefirst}}=\mathrm{\pm 1},\mathrm{\pm 2},\mathrm{\pm 4},\pm \frac{1}{2}$$

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

$$\begin{array}{l}\frac{p}{q}=\frac{\text{factorofconstantterm}}{\text{factorofleadingcoefficient}}\hfill \\ =\frac{\text{factorof1}}{\text{factorof2}}\hfill \end{array}$$

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

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

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

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.

$$\begin{array}{l}\frac{p}{q}=\frac{\text{factorofconstantterm}}{\text{factorofleadingcoefficient}}\hfill \\ =\frac{\text{factorof–1}}{\text{factorof4}}\hfill \end{array}$$

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.

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.

$$\begin{array}{l}\frac{p}{q}=\frac{\text{factorofconstantterm}}{\text{factorofleadingcoefficient}}\hfill \\ =\frac{\text{factorof3}}{\text{factorof3}}\hfill \end{array}$$

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 $\pm \frac{i\sqrt{3}}{3}.$

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.

$$\begin{array}{l}f(x)=a(x+3)(x-2)(x-i)(x+i)\\ f(x)=a(x^2+x-6)(x^2+1)\\ f(x)=a(x^4+x^3-5x^2+x-6)\end{array}$$

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

$$\begin{array}{l}100=a({(-2)}^4+{(-2)}^3-5{(-2)}^2+(-2)-6)\hfill \\ 100=a(-20)\hfill \\ -5=a\hfill \end{array}$$

So the polynomial function is

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

or

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

Begin by determining the number of sign changes.

Figure 3

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.

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

Figure 4

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.

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 $h=\frac{1}{3}w.$ Let’s write the volume of the cake in terms of width of the cake.

$$\begin{array}{l}V=(w+4)(w)(\frac{1}{3}w)\\ V=\frac{1}{3}{w}^3+\frac{4}{3}{w}^2\end{array}$$

Substitute the given volume into this equation.

$$\begin{array}{ll}351=\frac{1}{3}{w}^3+\frac{4}{3}{w}^2\hfill & \text{Substitute351for}V.\hfill \\ 1053=w^3+4w^2\hfill & \text{Multiplybothsidesby3}.\hfill \\ 0=w^3+4w^2-1053\hfill & \text{Subtract1053frombothsides}.\hfill \end{array}$$

Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are $\pm 1,\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351,$ and $\pm 1053.$ 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. Use synthetic division to check $x=1.$

$$\begin{array}{l}\hfill \\ \begin{array}{c}1\\ \end{array}\overline{)\begin{array}{rrrr}\hfill 1& \hfill 4& \hfill 0& \hfill -1053\\ \hfill & \hfill 1& \hfill 5& \hfill 5\end{array}}\hfill \\ \begin{array}{rrrr}\hfill \text{1}& \hfill 5& \hfill 5& \hfill -1048\end{array}\hfill \end{array}$$

Since 1 is not a solution, we will 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 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.

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

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