# Using the remainder theorem to find a remainder.

## Question

The polynomial ${\color{Blue}&space;x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ is divided by ${\color{Blue}&space;x+1}$. What is the remainder?

## Solution

Do you want to try this problem using polynomial long division? That could be a nightmare. How about synthetic division? I didn’t think so. You need the remainder theorem.

The remainder theorem states that when a polynomial (like, for example, the one in this problem) is divided by a binomial ${\color{Blue}&space;x-c}$, the remainder is equal to the value of the polynomial at point ${\color{Blue}&space;x=c}$.

Consider a simple example: When polynomial ${\color{Blue}&space;x^2-2x+2}$ is divided by ${\color{Blue}&space;x-1}$, the remainder is equal to the value of ${\color{Blue}&space;x^2-2x+2}$ at point ${\color{Blue}&space;x=1}$. Since 1 is an easy number at which to evaluate a function, it’s not too hard to see that at ${\color{Blue}&space;x=1}$, ${\color{Blue}&space;x^2-2x+2}=1.$  Thus when ${\color{Blue}&space;x^2-2x+2}$ is divided by ${\color{Blue}&space;x-1}$, the remainder is $1.$

You can confirm this result with polynomial long division or synthetic division.

Now try the question. To find the remainder when ${\color{Blue}&space;x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ is divided by ${\color{Blue}&space;x+1}$, evaluate ${\color{Blue}&space;x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$  at ${\color{Blue}&space;x=-1.}$ That’s

${\color{Blue}&space;(-1)^{55}-6&space;\cdot&space;(-1)^{42}-6&space;\cdot&space;(-1)^{29}-3&space;\cdot&space;(-1)^{23}&space;+&space;(-1)^4-2}$

Note that -1 taken to an even power is 1 and -1 taken to an even power is 1. Thus ${\color{Blue}&space;(-1)^{55}=-1}$, ${\color{Blue}&space;(-1)^{29}=-1}$, and ${\color{Blue}&space;(-1)^{23}=-1}$; while ${\color{Blue}&space;(-1)^{42}=1}$ and ${\color{Blue}&space;(-1)^{4}=1}$. The whole polynomial equals

${\color{Blue}&space;-1-\left&space;(6&space;\cdot&space;1&space;\right&space;)-\left&space;(6&space;\cdot&space;-1&space;\right&space;)-\left&space;(3&space;\cdot&space;-1&space;\right&space;)+1-2}$

${\color{Blue}&space;-1-6+6+3+1-2=1}$

The remainder is 1.