Evaluate $P(x) \bmod (x^2-x-2)$ 
Let $P(x)$ be a polynomial.
If $P(x) \bmod (x+1)=0$ and $P(x) \bmod (x-2)=6$, then evaluate $P(x) \bmod (x^2-x-2)$.

My attempt:
$$P(x)=k(x)(x-2)+6=k(x)(x+1)-3(k(x)-2)$$
$$\Longrightarrow (k(x))-2) \bmod (x+1)=0$$
$$\Longrightarrow \left(\dfrac{P(x)-6}{x-2}-2\right) \bmod (x+1)=0$$
$$\Longrightarrow (P(x)-2x-2) \bmod ((x+1)(x-2)) =0$$
$$\Longrightarrow P(x) \bmod (x^2-x-2) \equiv (2x+2) \bmod ((x+1)(x-2)) = (2x+2) \bmod (x^2-x-2)=2x+2 $$
Is my solution correct?

In as simple a case as this we can just proceed this way.
We have $P(x)= ax+b \mod (x^2-x-2)$ for some $a,b$.
That is $P(x)=ax+b+ q(x)(x+1)(x-2)$.
Putting $x=-1$ and $x=2$ gives us $-a+b=0$ and $2a+b=6$.
Hence $a=b=2$.
But in general I'd suggest that the Extended Chinese Remainder Algorithm is the way to go.
