Find all integers $k$ such that $\frac{k^4+k^3-5k^2+k}{k^2+k-6}$ is an integer. Find all integers $k$ such that $\frac{k^4+k^3-5k^2+k}{k^2+k-6}$ is an integer. 
I tried factorization and ended up with $\frac{k(k^3+k^2-5k+1)}{(k-2)(k+3)}$. Please give me some hints. Thank you!

HINT:
$$\frac{k^4+k^3-5k^2+k}{k^2+k-6}=\frac{(k^2+1)(k^2+k-6)+6}{k^2+k-6}$$
