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!
 A: Just to add an alternative: From an algorithmic point of view (i.e., to write a program that performs factorization without any intelligence guiding the process), one of the fastest things to check and therefore typically the first step is squarefree factorization, by taking gcds of the polynomial and its derivative(s).
$$\begin{eqnarray*} f&=&n^4+6n^3+11n^2+6n+1\\ f'&=& 4n^3+18n^2+22+6\\ \gcd(f,f')&=& n^2+3n+1 \end{eqnarray*}$$
Of course, working by hand and with a human brain behind it, the cost analysis may be different from that for a computer and computing gcds may not be as attractive as it is for machines.
A: 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 $.
A: we can reorganize the terms of $n^4+6n^3+11n^2+6n+1$ like $$(n^4+2n^2+1)+(6n^3+6n)+9n^2 $$ $$(n^2+1)^2+6n(n^2+1)+(3n)^2=(n^2+3n+1)^2$$ then we are done
A: I was thinking along the same lines as TZakrevskiy but with a slight twist.  Once you've become convinced that $P(n)=n^4+6n^3+11n^2+6n+1$ is the square of a quadratic (because its values for small $n$ are all squares), try computing
$$P(10)=10000+6000+1100+60+1=17161$$
and then take its square root:
$$\sqrt{17161}=131=1\cdot10^2+3\cdot10+1$$
which suggests the (correct) factorization
$$P(n)=(n^2+3n+1)^2$$
If you want to be extra sure, look at
$$\sqrt{100^4+6\cdot100^3+11\cdot100^2+6\cdot100+1}=10301$$
A: In general there's no easy way.  But there are a lot of tricks you can use, and sometimes they work.
For this problem, the rational root theorem tells us that if there is a rational root, it must be $\pm 1$.  Evidently it isn't either of these, so there are no rational roots, and therefore no factors of the form $n-q$.
So we are looking for two second-degree factors: $$n^4+6n^3+11n^2+6n+1 = (an^2+bn+c)(pn^2+qn+r)$$
But we can see from the coefficient of 1 on the $n^4$ term that $a=p=1$. From the constant term of 1 we have either  $c=r=1$ or $c=r=-1$, so it's really 
$$n^4+6n^3+11n^2+6n+1 = (n^2+bn\pm1)(n^2+qn\pm1).$$
Equating the  coefficients of the $n^3$ terms on each side, we see that $b+q=6$.
Equating the coefficients of the $n^2$ terms on each side, we see that either $bq+2=11$ (if we take the $\pm$ signs to be $+$) or that $bq-2 = 11$ (if the $\pm$ signs are $-$).
The equations $b+q=6$ and $bq-2 = 11$ certainly have no solution in integers, since one of $b$ or $q$ must be 13.  So we try the other pair of equations, $b+q=6$ and $bq+2 = 11$.  Here the solution $b=q=3$ is obvious, and we are done, except to check the answer.  We must do this, because we have completely ignored  the coefficient of the $n$ term.
A: 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 $k^2 - 2 = n^2 + \frac{1}{n^2}$.
