Some polynomial problems

[earthboundslave]

When a polynomial f(x) is divided by (x-2), the remainder is 7. When f(x) is divided by (x+1) the remainder is -2.

(a) If the remainder is px+q when f(x) is divided by (x-2)(x+1), find the values of p and q.

(b) Find the remainder when f(x+3) is divided by (x+1)(x+4)

I found that p=3 and q=1 using the remainder theorem. f(x)=Q(x)*(x-2)(x+1)+(px+q)

However I don't know how to get the remainder of part(b).

Let g(x)=f(x+3).
When g(x) is divided by (x+1), the remainder
=g(-1)
=f(-1+3)
=f(2)

When g(x) is divided by (x+4), the remainder
=g(-4)
=f(-4+3)
=f(-1)

At this point, since the two remainders I got are the same when f(x) is divided by (x-2) and (x+1), I assume f(x+3)=Q(x)*(x+1)(x+4)+(px+q).
But 3x+1 is not the answer.
What mistakes have I made and how to solve part(b)?

 

[MarkFL]

We know from the given information that:

[MATH]f(-1)=-2[/MATH]
[MATH]f(2)=7[/MATH]
And thus:

[MATH]f(-1)=p(-1)+q=-2\implies p-q=2[/MATH]
[MATH]f(2)=p(2)+q=7\implies 2p+q=7[/MATH]
And this implies:

[MATH](p,q)=(3,1)\quad\checkmark[/MATH]
For part b), I would let \(x=u-3\), and so we are simply being asked for the remainder when \(f(u)\) is divided by \((u-2)(u+1)\), and we found from part a), that this must be:

[MATH]3u+1[/MATH]
So, what do we get when we back-substitute for \(u\)?