Factoring a hard polynomial This might seem like a basic question but I want a systematic way to factor the following polynomial:
$$n^4+6n^3+11n^2+6n+1.$$
I know the answer but I am having a difficult time factoring this polynomial properly. (It should be $(n^2 + 3n + 1)^2$). Thank you and have a great day!

There's another approach: check that for, say $n=\pm 1,\, 0,\,\pm 2$ your polynomial is a perfect square:$$\begin{cases}25,&n=1\\1,&n=-1\\1,&n=0\\1,&n=-2\\121,&n=2.\end{cases}$$ Therefore,  in five points a polynomial of fourth degree is a perfect square - you can hope to find a polynomial of the second degree that satisfies
$$\begin{cases}\pm 5,&n=1\\\pm 1,&n=-1\\\pm 1,&n=0\\\pm 1,&n=-2\\\pm 11,&n=2\end{cases}.$$
This is an overdetermined system, but, luckily, it has a solution: wlog, we take the candidate as $ n^2+2bn+c$ (because of the coefficient at $n^4$). The value in $n=-1$ gives $1-2b+c = \pm 1$, and in $n=0$ $ c=\pm 1$. By checking $4$ solutions of the obtained $4$ linear systems we can eliminate all but one by testing against remaining points $n$: we obtain $c=1$, $b=3/2$. Finally we check that indeed $(n^2+3n+1)^2=n^4+6n^3+11n^2+6n+1 $.
