Proving $n(n-2)(n-1)^2$ is divisible by 12 I want to show that $n^4 - 4n^3 + 5n^2 -2n$ is divisible by $12$ whenever $n>0$.
I reduced this to $n(n-2)(n-1)^2$. My approach has been to check divisibility by $3$ and $4$.
In both cases, squares are always congruent to $0$ or $1$. So
\begin{align*}n^2 &\equiv_3 0, 1 \\ \implies (n-1)^2 &\equiv_3 n-2, n-1\end{align*}
Given $-1\equiv_3 2$, then $(n-1)^2\equiv_3 n-2$ or $(n-1)^2\equiv_3 n-1$. But now I'm stuck here.
I know the polynomial is made up of a product of 3 consecutive numbers, so either two of them are even and one odd, or two of them are odd and one even. When I made the assumption $n$ is even, then $n-1$ is odd and $n-2$ is even, eventually landing at $n(n-2)(n-1)^2 \equiv_4 n(n-2)$ but so far nothing helpful has come up.
What am I missing?

Suppose $(n,12)=1.$  We use Euler and lil' Fermat.
$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 2(n-1)\pmod {2^2},$$
and
$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 0\pmod 3.$$
So by CRT, using Bezout,  we have $1×4×2×0-1×3×2(n-1)\equiv6(n-1)\equiv0\pmod{12}.$
Suppose $(n,12)\neq1.$  Then either $2\mid n$ or $3\mid n.$
If $2\mid n,$ then if $3\not \mid n$, we are done by the previous calculation and CCRT.
If $3\mid n,$ then $n^4-4n^3+5n^2-2n\equiv 0\pmod3.$  So we're done by CCRT.
If $2\not\mid n$, then $3\mid n,$ so we're back to Bezout,  since we have already done the modular arithmetic: $$1×4×0-1×3×2(n-1)\equiv 0\pmod {12}.$$
