6.3.4 Dividing Polynomial Functions and the Remainder Theorem

Dividing Polynomial Functions

Just as polynomials can be divided, polynomial functions can also be divided.

DIVISION OF POLYNOMIAL FUNCTIONS

For functions $f(x)$ and $g(x)$, where $g(x)\ne 0$,

$(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$

Example 1

For functions $f(x)=x^2-5x-14$ and $g(x)=x+2$:

a. Find $(\frac{f}{g})(x)$.

Step 1 - $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$ Substitute for $f(x)$ and $g(x)$.

$(\frac{f}{g})(x)=\frac{f(x)=x^2-5x-14}{x+2}$

Step 2 - Divide the polynomials using long division.

$$\begin{array}{l}x-7\\ x+2)\phantom{1x^2-5x-14}\overline{\phantom{1}{x}^2-5x-14}\\ \underset{\_}{-(x^2+2x)}\\ -7x-14\\ \underset{\_}{-(-7x-14)}\\ 0\end{array}$$

$(\frac{f}{g})(x)=x-7$

b. Find $(\frac{f}{g})(-4)$.

Step 1 - In part (1), we found $(\frac{f}{g})(x)$.

$(\frac{f}{g})(x)=x-7$

Step 2 - To find $(\frac{f}{g})(-4)$, substitute $x=-4$.

$(\frac{f}{g})(-4)=-4-7$

$(\frac{f}{g})(-4)=-11$

Using the Remainder Theorem

Let’s look at some division problems and their remainders. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as $x-c$, the value of the function at $c$, $f(c)$, is the same as the remainder from the division problem.

To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and adding the remainder. In function notation, we could say: To get the dividend $f(x)$, we multiply the quotient, $q(x)$, times the divisor, $x-c$, and add the remainder, $r$.

$f(x)=q(x)(x-c)+r$

If we evaluate this at $c$, we get:

$f(c)=q(c)(c-c)+r$

$f(c)=q(c)(0)+r$

$f(c)=r$

This leads us to the Remainder Theorem.

Example 2

Use the Remainder Theorem to find the remainder when $f(x)=x^3+3x+19$ is divided by $x+2$.

To use the Remainder Theorem, we must use the divisor in the $x-c$ form. We can write the divisor $x+2$ as $x-(-2)$. So, our $c$ is $-2$.

To find the remainder, we evaluate $f(c)$, which is $f(-2)$.

$f(x)=x^3+3x+19$

Step 1 - To evaluate $f(-2)$, substitute $x=-2$.

$f(-2)=(-2{)}^3+3(-2)+19$

Step 2 - Simplify.

$f(-2)=(-8)-6+19$

$f(-2)=5$

The remainder is 5 when $f(x)=x^3+3x+19$ is divided by $x+2$.

Step 3 - Check using synthetic division.

The remainder is 5.