SOLUTION: Here is my question: 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 remainder when P(X) is divided by (x-1)(x-2)(x-3).

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Here is my question: 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 remainder when P(X) is divided by (x-1)(x-2)(x-3).       Log On


   

Question 758299: Here is my question:
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 remainder when P(X) is divided by (x-1)(x-2)(x-3).
here is part of my work:
let the quotient be Q(x), and let the remainder be ax+b
P(x)=(x-1)(x-2)(x-3)Q(X)+ax+b
P(1)=a*1+b=1 then a+b=1 (1)
then I don't know how to do with (x-2)(x-3) and the remainder is 5. Please help!
thanks
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-1%29%28x-2%29%28x-3%29 is cubic (degree=3) polynomial.
When you divide by %28x-1%29%28x-2%29%28x-3%29 the remainder polynomial could have degree 2, 1, or 0, so you should start with ax%5E2%2Bbx%2Bc for a remainder.

P%28x%29=%28x-1%29%28x-2%29%28x-3%29Q%28X%29%2Bax%5E2%2Bbx%2Bc
P%281%29=a%2A1%5E2%2Bb%2A1%2Bc=a%2Bb%2Bc=1
P%282%29=a%2A2%5E2%2Bb%2A2%2Bc=4a%2B2b%2Bc=5
P%283%29=a%2A3%5E2%2Bb%2A3%2Bc=9a%2B3b%2Bc=5

That gives you the sytem
system%28a%2Bb%2Bc=1%2C4a%2B2b%2Bc=5%2C9a%2B3b%2Bc=5%29
system%28a%2Bb%2Bc=1%2C4a%2B2b%2Bc=5%29 --> 3a%2Bb=4 and
system%28a%2Bb%2Bc=1%2C9a%2B3b%2Bc=5%29 --> 8a%2B2b=4, so
system%28a%2Bb%2Bc=1%2C4a%2B2b%2Bc=5%2C9a%2B3b%2Bc=5%29 --> system%28a%2Bb%2Bc=1%2C3a%2Bb=4%2C8a%2B2b=4%29
Then,
system%283a%2Bb=4%2C8a%2B2b=4%29 --> 8a%2B2b-6a-2b=4-8%29 --> 2a=-4%29 --> highlight%28a=-2%29
system%283a%2Bb=4%2Ca=-2%29 --> -6%2Bb=4 --> highlight%28b=10%29
and finally
system%28a%2Bb%2Bc=1%2Ca=-2%2Cb=10%29 --> -2%2B10%2Bc=1 --> 8%2Bc=1 --> highlight%28c=-7%29
The remainder when P%28x%29 is divided by %28x-1%29%28x-2%29%28x-3%29 is
highlight%28-2x%5E2%2B10x-7%29

There are infinite possibilities for P%28x%29, as there are for Q%28x%29. simplest P%28x%29 corresponds to Q%28x%29=1
In that case,