Problem with a product of 2 remainders (polynomials)

[another_dude]

Homework Statement

[/B]
Polynomial P(x) when divided by (x-2) gives a remainder of 10. Same polynomial when divided by (x+3) gives a remainder of 5. Find the remainder the polynomial gives when divided by (x-2)(x+3).

2. Homework Equations

Polynomial division, remainder theorem

The Attempt at a Solution

From the problem and the remainder theorem I got 2 equations:
1) P(2)=10 and 2) P(-3)=5
More generally, we get this: P(x)=(x-2)D(x)+10=(x+3)F(x)+5 , where D(x), F(x) some polynomials one rank lower than P(x).

Tried multiplicating, dividing and adding the equations in different ways but couldn't get a useful relation, namely P(x)=(x-2)(x+3)G(x)+ V(x), where V(x) is the wanted remainder. Any thoughts?

 

[fresh_42]

Homework Statement

[/B]
Polynomial P(x) when divided by (x-2) gives a remainder of 10. Same polynomial when divided by (x+3) gives a remainder of 5. Find the remainder the polynomial gives when divided by (x-2)(x+3).

2. Homework Equations

Polynomial division, remainder theorem

The Attempt at a Solution

From the problem and the remainder theorem I got 2 equations:
1) P(2)=10 and 2) P(-3)=5
More generally, we get this: P(x)=(x-2)D(x)+10=(x+3)F(x)+5 , where D(x), F(x) some polynomials one rank lower than P(x).

Tried multiplicating, dividing and adding the equations in different ways but couldn't get a useful relation, namely P(x)=(x-2)(x+3)G(x)+ V(x), where V(x) is the wanted remainder. Any thoughts?

Do you know the Chinese remainder theorem?

 

[another_dude]

Never heard of it before.

 

[Dick]

Tried multiplicating, dividing and adding the equations in different ways but couldn't get a useful relation, namely P(x)=(x-2)(x+3)G(x)+ V(x), where V(x) is the wanted remainder. Any thoughts?

$V(x)$ must have the form $ax+b$, right? What are $V(2)$ and $V(-3)$?

 

[another_dude]

Oh right, didn't think of that actually. Well, from the remainder theorem we get 1) V(2)=P(2)=10 2) V(-3)=P(-3)=5 . Then you solve the system for a and b. Thanks!