# Math Help - Remainders of a Unknown Polynomial.

1. ## 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.

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

3. 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.

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

5. $
R(x)=ax^2+bx+c
$

$
R(x)=(x-2)(x-3)a +(b+5a)x+(c-6a)
$

if my polynomial long division is correct. I'm a bit rusty. so that would mean
$b+5a = 0$ and $c-6a = 5$?

thanks again for all your help.