Divisibility of Polynomials: Finding the Remainder

[Hannisch]

Homework Statement

A polynomial p(x) leaves the rest 3 when divided by (x+2) and the rest 8 when divided by (x-6). What's the rest r(x) when p(x) is divided by (x+2)(x-6)?

Homework Equations

The Attempt at a Solution

I wrote the three equations:

p(x)=q₁(x+2) + 3
p(x)=q₂(x-6) + 8
p(x)=q₃(x+2)(x-6) + r(x)

And I've tried rearranging them and I've tried to find what p(x) is and I've tried some other random things and it doesn't work out. I think the main problem is that I don't know at which angle to approach this problem and that I don't know how to reason about it..

 

[tiny-tim]

Hi Hannisch!

p(x)=q₁(x+2) + 3
p(x)=q₂(x-6) + 8
p(x)=q₃(x+2)(x-6) + r(x)

Hint: now use the fact that r(x) must be of the form Ax + B

 

[Dick]

Multiply your first equation by (x-6) and your second equation by (x+2) and subtract them.

 

[Hannisch]

I can get an answer using Dick's method:

p(x)=q₃(x+2)(x-6)+ (5/8)x + (34/8), where q₃ = (q₂-q₁)/8

But I don't know if it's correct or not? (I think it is, because I've got another problem just like this one that I do have the answers to and I applied the same method and got the correct answer.)

And I'm not sure how to apply tiny-tim's tip? I mean, sure I can put p(x)=q³(x+2)(x-6) + r(x) = p(x)=q₃(x+2)(x-6) + ax + b, and then again try to fix it, but basically I fall back into the same stupid situation I was in before..

 

[Dick]

tiny-tim is just suggesting you use p(-2)=3=r(-2) and p(6)=8=r(6). If you set r(x)=ax+b, that's two equations in the two unknowns, a and b. And your answer is just fine. Check it with this method.

 

[Hannisch]

Ohh, I get it now (it took me a while, I've really stared myself blind on this problem)! Anyway, I do get the same answer now (I got it wrong first, but it was a ridiculous mistake, really), so thank you very, very much!