# [SOLVED] Polynomial division

• Aug 28th 2007, 12:21 PM
Spec
[SOLVED] Polynomial division
Not really a homework question, but I have a test tomorrow and I need to be able to solve these kind of problems.

Find the remainder when the polynomial \$\displaystyle x^{100} + x^{67} - x^{32} - 2x^9 +1\$ is divided with:

a) \$\displaystyle x - 1\$
b) \$\displaystyle x + 2\$
c) \$\displaystyle x^2 - x\$

I already have the answers, what I'm looking for is how to solve the problems. Polynomial long division would easily fill up several pages, so I'm clueless as what to do.

• Aug 28th 2007, 01:40 PM
ThePerfectHacker
Quote:

Originally Posted by Spec
Not really a homework question, but I have a test tomorrow and I need to be able to solve these kind of problems.

Find the remainder when the polynomial \$\displaystyle x^{100} + x^{67} - x^{32} - 2x^9 +1\$ is divided with:

a) \$\displaystyle x - 1\$
b) \$\displaystyle x + 2\$
c) \$\displaystyle x^2 - x\$
!

Theorem: Given \$\displaystyle f(x),g(x)\$ two polynomials over \$\displaystyle \mathbb{Z}\$ with \$\displaystyle g(x)\not = 0\$ then \$\displaystyle f(x) =q(x)g(x)+r(x)\$ and \$\displaystyle \deg r(x) < \deg g(x) \mbox{ or }r(x)=0\$.

So,
\$\displaystyle x^{100} + x^{67} - x^{32} - 2x^9 +1=q(x)(x-1)+k\$.
It is a constant since the degree is zero.
Since the above statement is true for all \$\displaystyle x\$ let \$\displaystyle x=1\$ to get:
\$\displaystyle 0=q(1)(0)+k\implies k=0\$.
Thus there is no remainder.

Try the same approach to the other ones.