Find $p$ if the remainder when $(x^2 + px + 13)$ divided by $(x-2)$ is twice the remainder when it is divided by $(x+2)$.

Question

As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.

$x^2+px+13$

With $(x-2)$

$f(2)=(2)^2+p(2)+13$

$f(2)=17+2p=2r$

With $(x+2)$

$f(-2)=(-2)^2+p(-2)+13$

$f(-2)=17-2p=r$

I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.

Answer 1

You are making it too hard.

$f(2) = 17+2p, f(-2) = 17-2p$.

So you want $17+2p = 2(17-2p) =34-4p$ so $17=6p$ or $p = 17/6$ as you got.

Answer 2

HINT:

$$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$

Via polynomial long division.