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!
 A: standard trick. We have $11 \equiv 2 \pmod 3$ from which follows that the nontrivial cube roots of unity are roots,
$  \omega = \frac{-1+i \sqrt3}{2} , \bar{\omega} = \omega^2 =  \frac{-1-i \sqrt3}{2}$
This means the original polynomial is divisible by
$$ (x-\omega) ( x - \bar{\omega} ) = \; \;  x^2 + x +1 $$
$$  \left(   x^{11}  +  x  + 1 \right)  =  \left(   x^{2}  +  x  + 1 \right)  \cdot \color{magenta}{  \left(   x^{9}  -  x^{8}  +  x^{6}  -  x^{5}  +  x^{3}  -  x^{2}  + 1 \right) }  $$
A: 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}
A: No tricks: Let $f=x^{11}+x+1$ and suppose $g,h\in\Bbb{Z}[x]$ are such that $f=gh$ and $\deg g=2$. Then
$$g=ax^2+bx+c,$$
for some integers $a,b,c\in\Bbb{Z}$. Then $f=gh$ shows that $a$ divides the leading term of $f$, and $c$ divides the constant term of $f$, so $a=\pm1$ and $c=\pm1$. Without loss of generality $a=1$, and so
$$g=x^2+bx\pm1.$$
Now if $\alpha$ is a root of $g$ then $g(\alpha)=0$ and $f(\alpha)=0$, meaning that
$$\alpha^2=-b\alpha\mp1\qquad\text{ and }\qquad\alpha^{11}+\alpha+1=0.$$
and from here it's a cumbersome but straightforward bit of algebra to determine $b$.
A: Now that I know the answer, thanks to @Will Jagy, here is my solution:
$$x^{11} + x^2 + 1 = (x^{11} - x^2) + (x^2 + x + 1) = \\
= x^2(x^9-1) + (x^2 + x + 1)$$
Note that the first term contains the factor $x^9-1$ which is divisible by $x^3-1= (x-1)(x^2+x+1)$.
Therefore, our polynomial is also divisible by $x^2 + x + 1$.
Generalization: If $a_1$,$\ldots$, $a_n$ are natural numbers that give different remainders when divided by $n$ then the polynomial
$$x^{a_1} + \cdots + x^{a_n}$$
is divisible by $x^{n-1} + \cdots + x + 1$.
Even more,  a polynomial $ \sum c_k x^k$ is divisible by $x^{n-1} + \cdots + x + 1$ if and only if we have
$$s_0 = s_1 = \cdots = s_{n-1}$$
where $$s_r = \sum_{k \equiv r \mod n} c_k$$
