# Remainders of a Unknown Polynomial.

• June 28th 2010, 06:31 PM
Remainders of a Unknown Polynomial.
When P(x) is divided by x-1 the remainder is 1, and when divided by (x-2)(x-3), the remainder is 5. Find the remainder when P(x) is divided by (x-1)(x-2)(x-3).

I'm using the formula where
P(x)=(x-1)(x-2)(x-3)Q(x)+ax^2+bx+c
--

hints / steps to solve this will be greatly appreciated. thanks in advance.
• June 28th 2010, 08:46 PM
CaptainBlack
Quote:

When P(x) is divided by x-1 the remainder is 1, and when divided by (x-2)(x-3), the remainder is 5. Find the remainder when P(x) is divided by (x-1)(x-2)(x-3).

I'm using the formula where
P(x)=(x-1)(x-2)(x-3)Q(x)+ax^2+bx+c
--

hints / steps to solve this will be greatly appreciated. thanks in advance.

Put $R(x)=ax^2+bx+c$ then $R(x)=(x-1)(ax+(a+b)) + (a+b+c)$, but we know that the remainder when $p(x)$ is divided by $x-1$ is $1$ so we have $a+b+c=1$.

A similar process for the other remainder will give you two more equations in $a,\ b$ and $c$ which you then solve.

CB
• June 28th 2010, 09:15 PM
Thank you, CB! It makes sense now. I just couldn't figure it out trying to help the one who showed me this problem. Although, he might have to re-iterate it to make sure it's clear.

From my understanding of the problem, however. there are only two remainders given. one when dividing by (x-1) and the other by (x-2)(x-3), but in that case you would have one dependent variable. and therefore many remainders.
• June 28th 2010, 10:40 PM
CaptainBlack
Quote:

Thank you, CB! It makes sense now. I just couldn't figure it out trying to help the one who showed me this problem. Although, he might have to re-iterate it to make sure it's clear.

From my understanding of the problem, however. there are only two remainders given. one when dividing by (x-1) and the other by (x-2)(x-3), but in that case you would have one dependent variable. and therefore many remainders.

For the other remainder you have:

$ax^2+bx+c=a(x-2)(x-3)+5$

Expand and equate coefficients and you will find that you have two equations.

CB
• June 30th 2010, 06:07 PM
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$b+5a = 0$ and $c-6a = 5$?