Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$ 
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgusting, I was mixing up in calculations and couldn't finally get answer.
So, Is there any better way to find such a polinomial (divisor of this one with degree 2)? It would be awesome if proposed way would not require super advanced knowledge. Thanks in advance!

Factorize as follows
\begin{align}x^{11} + x + 1
&= \frac{ x^{11}(x-1) + (x+1)(x-1)}{x-1}\\
&= \frac{x^{12}-x^{11}+x^2-1}{ x-1}
= \frac{(x^{12}-1 )-x^2(x^{9}-1)}{ x-1}\\
&= \frac{(x^{3}-1)(x^9+x^6+x^3+1)-x^2(x^{3}-1)(x^6+x^3+1)}{ x-1}\\
&= \frac{x^{3}-1}{x-1}(x^9+x^6+x^3+1-x^{8}-x^5-x^2)\\
 &= (x^2+x+1)(x^9-x^8+x^6-x^{5}+x^3-x^2+1)
\end{align}
To generalize, follow the same steps above
\begin{align}
x^5+x+1 &= (x^2+x+1)(x^3-x^2+1)\\
x^{17}+x+1 &= (x^2+x+1)(x^{15}-...-x^2+1)\\
...&= \> ...\\
x^{6n-1}+x+1 &= (x^2+x+1)(x^{6n-3}-...-x^2+1)\\
\end{align}
