Factorise $x^5+x+1$ 
Factorise $$x^5+x+1$$

I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?

If you suspect there exists a factorization over $\mathbb{Q}[x]$ into polynomials of degree 2 and 3, but you just don't know their coefficients, one way is to write it down as
$x^5 + x + 1 = (x^2 + ax + b)(x^3 + cx^2 + dx + e)$ where the coefficients are integers (by Gauss' lemma). And then expand and solve the system.
Then $a + c = 0, b + ac + d = 0, bc + ad + e = 0, ae + bd = 1, be = 1$.
So we get $c = -a$ and $b = e = 1$ or $b = e = -1$.
In the first case we reduce to $1 - a^2 + d = 0, -a + ad + 1 = 0, a + d = 1$ which gives $d = 1 - a, 1 - a^2 + 1 - a = 0, 1 - a^2 = 0$ so $a = 1, b = 1, c = -1, d = 0, e = 1$.
Substituting gives us $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$
If the factorization was not over $\mathbb{Q}[x]$ then things would get more complicated because I could not assume $b = e = +/- 1$.
