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!

Exploit the symmetry of the palindromic polynomial to factorise as follows,
$$\begin{align*}
n^4 + 6n^3 + 11n^2 + 6n + 1
&= n^2\left( n^2 + \frac{1}{n^2} + 6n + \frac{6}{n} + 11 \right)\\
&= n^2\left( (n + 1/n)^2 + 6(n + 1/n) + 9 \vphantom{\frac{1}{n^2}} \right)\\
&= n^2\left( n + 1/n + 3 \right)^2\\
&= \left( n^2 + 3n + 1 \right)^2,\\
\end{align*}$$
having essentially substituted $k = n + \frac{1}{n}$ and used $k^2 - 2 = n^2 + \frac{1}{n^2}$.
