Using the remainder theorem to find a remainder.

Question

The polynomial ${\color{Blue} x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ is divided by ${\color{Blue} x+1}$ . 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 ${\color{Blue} x-c}$ , the remainder is equal to the value of the polynomial at point ${\color{Blue} x=c}$ .

Consider a simple example: When polynomial ${\color{Blue} x^2-2x+2}$ is divided by ${\color{Blue} x-1}$ , the remainder is equal to the value of ${\color{Blue} x^2-2x+2}$ at point ${\color{Blue} x=1}$ . Since 1 is an easy number at which to evaluate a function, it’s not too hard to see that at ${\color{Blue} x=1}$ , ${\color{Blue} x^2-2x+2}=1.$ Thus when ${\color{Blue} x^2-2x+2}$ is divided by ${\color{Blue} x-1}$ , 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 ${\color{Blue} x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ is divided by ${\color{Blue} x+1}$ , evaluate ${\color{Blue} x^{55}-6x^{42}-6x^{29}-3x^{23}+x^4-2}$ at ${\color{Blue} x=-1.}$ That’s

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

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