What is the sum of all integers $n$ such that $n^2 + 2n +2$ divides $n^3 + 4n^2 + 4n -14$? How to proceed in this problem? I divided  $n^3 + 4n^2 + 4n -14$ by $n^2 + 2n +2$ and equated the remainder to zero . I got $n=5$ as answer but the answer is wrong. 

It must help:
$$\frac{n^3+4n^2+4n-14}{n^2+2n+2}=n+2-\frac{2n+18}{n^2+2n+2}.$$
Thus, or
$$\left|\frac{2n+18}{n^2+2n+2}\right|\geq1,$$ which gives
$$-4\leq n\leq4$$
I got $n\in\{0,1,4,-2,-4\}$.
Also, we have $n=-9$.
