Find all $n\ge2$ such that $x^2+2$ divides $x^5-10x+12$ in $\mathbb{Z}_n$.
To begin, I divided $x^5-10x+12$ by $x^2+2$ which gave me: $$x^5-10x+12 = (x^3-2x)(x^2+2)-6x+12$$
So, I guess I need to find $\mathbb{Z}_n$ in a way that $-6x+12$ is congruent to $0$, right? But I have one question before it. Can I just care about $-6x+12$ and don't care when, for example, $x^2+2$ will be congruent to $0$ in some $\mathbb{Z}_k$? For example, if I pick $\mathbb{Z}_6$ I get what I desire, but what about $\mathbb{Z}_2$? I wouldn't be dividing by $x^2+2$ anymore because it is $x^2$ in $\mathbb{Z}_2$. Also, will the answer be every $\mathbb{Z}_{2k}$?
In $\mathbb{Z}_2$ we have $-6x+12=0$, so indeed $x^2+2$ divides $x^5-10x+12$. The fact that $x^2+2=x^2$ is more or less irrelevant. The polynomial $x^2+2$ is monic, so your division is correct in all $\mathbb{Z}_n[x]$ with $n\ge 1$.
The answer will not be all $\mathbb{Z}_{2k}$. For example, $-6x+12$ is not the zero polynomial in $\mathbb{Z}_{14}$, and indeed is already non-zero in $\mathbb{Z}_8$.