$f(x)= x^2 -6x -9$ and $f(p)=f(-q)$. Find $p-q$

Question

The remainder when $x^2-6x-9$ is divided by $x-p$ is the same as when it is divided by $x+q$ where $p\ne-q$. Find the value of $p-q$.

Could you please explain how and why your method worked?

I tried multiplying $(x-p)(x+q)$ but my teacher told me the method was wrong, even though my answer was correct. Could you please explain why it is wrong, if wrong; why correct, if correct.

Answer 1

Taylor $\,\Rightarrow\,\dfrac{f(x)-f(p)}{x-p} = f'(p) + f''(p)(x-p)/2\ $

So $\,x=-q\,\Rightarrow 0 = 2p-6 -q - p \,\Rightarrow\, p-q = 6$