# Finding remainder of polynomial division

• Oct 29th 2008, 09:46 AM
mwok
Finding remainder of polynomial division
Find the remainder if the polynomial f(x) = 3x^100 + 5x^87 - 4x^40 + 2x^21 - 6 divided by p(x) = x + 1

The thing is, I can't use synthetic OR long division BECAUSE I'd have a VERY VERY long equation (since you have to fill in the missing degrees: 99,98,97,96,95......etc).

How would you solve this?
• Oct 29th 2008, 09:50 AM
masters
Quote:

Originally Posted by mwok
Find the remainder if the polynomial f(x) = 3x^100 + 5x^87 - 4x^40 + 2x^21 - 6 divided by p(x) = x + 1

The thing is, I can't use synthetic OR long division BECAUSE I'd have a VERY VERY long equation (since you have to fill in the missing degrees: 99,98,97,96,95......etc).

How would you solve this?

Use the Remainder theorem and find f(-1).
• Oct 29th 2008, 09:54 AM
mwok
According to The Remainder Theorem

How exactly would I use the remainder theorem????
• Oct 29th 2008, 10:17 AM
masters
Quote:

Originally Posted by mwok
According to The Remainder Theorem

How exactly would I use the remainder theorem????

The remainder theorem: When a polynomial P(x) is divided by x - a , the remainder is P(a)!

Your divisor is (x + 1). This means a = -1. Now, find P(a)...which in your case is P(-1).

$\displaystyle P(-1)=3(-1)^{100} + 5(-1)^{87} - 4(-1)^{40} + 2(-1)^{21} - 6$

This should yield the remainder you seek.
• Oct 29th 2008, 10:23 AM
Black Kawairothlite
you should use it at this way:

$\displaystyle \frac{{3x^{100}+5x^{87}-4x^{40}+2x^{21}-6}}{{x+1}}$

so let's say that:

$\displaystyle p(x)=3x^{100}+5x^{87}-4x^{40}+2x^{21}-6$

and $\displaystyle q(x)$ is the unknown answer

so rewriting the cocient and according to the remainder theorem:

$\displaystyle p(x)=(x+1)q(x)+r(x)$ where r(x) is the remainder

then:

$\displaystyle p(-1)=(0)q(-1)+r(-1)$

so $\displaystyle p(-1)=r(-1)$

$\displaystyle 3(-1)^{100}+5(-1)^{87}-4(-1)^{40}+2(-1)^{21}-6=r(-1)$

knowing the remainder then use synthetic division to do the operation which is easier to do.
• Oct 29th 2008, 10:26 AM
mwok
I finally got it, thanks!
• Oct 29th 2008, 10:31 AM
mwok
Quote:

Originally Posted by Black Kawairothlite
you should use it at this way:

$\displaystyle \frac{{3x^{100}+5x^{87}-4x^{40}+2x^{21}-6}}{{x+1}}$

so let's say that:

$\displaystyle p(x)=3x^{100}+5x^{87}-4x^{40}+2x^{21}-6$

and $\displaystyle q(x)$ is the unknown answer

so rewriting the cocient and according to the remainder theorem:

$\displaystyle p(x)=(x+1)q(x)+r(x)$ where r(x) is the remainder

then:

$\displaystyle p(-1)=(0)q(-1)+r(-1)$

so $\displaystyle p(-1)=r(-1)$

$\displaystyle 3(-1)^{100}+5(-1)^{87}-4(-1)^{40}+2(-1)^{21}-6=r(-1)$

knowing the remainder then use synthetic division to do the operation which is easier to do.

Book states f(x) = p(x) * q(x) + r(x) r(x) being remainder

where did you get the zero in (0)q(-1)?
• Oct 29th 2008, 02:51 PM
Black Kawairothlite
when you replace x @ (x+1) to -1 then (-1+1)=(0)

btw i saw that on the link u put there