TITLE: How can I show that the polynomial $p = x^5 - x^3 - 2x^2 - 2x - 1$ is irreducible over $\Bbb Q$?
QUESTION [8 upvotes]: I've tried a few "criteria" to check if this is irreducible. According to Maple it only has one entirely real root which I suspect is not rational but I can't prove it so I'm attempting to check if $p$ is irreducible. 
Eisenstein's Criterion doesn't work here and I'm yet to find a suitable transformation such that it could work. I also read that if a polynomial is irreducible over $\Bbb F_q$, with $q$ a prime not dividing the leading coefficient, then it is irreducible over $\Bbb Q$ so I reduced the polynomial modulo $2$ to obtain
$$p \equiv x^5 + x^3 + 1 \mod 2.$$
I think this is correct but then I need to know how to check the irreducibility of this new polynomial over $\Bbb F_2$. Do I simply need to check that neither $0$ nor $1$ are roots of this polynomial? (And am I applying this theorem correctly?)
If this polynomial IS irreducible over $\Bbb Q$, is the splitting field obtained by simply adjoining the roots to $\Bbb Q$?

REPLY [9 votes]: Your proof is almost complete: it is indeed sufficient to prove that $p(x)\in \mathbb F_2[x]$ (the reduction mod $2$ of your polynomial) is irreducible.
The polynomial $p(x)$ has no factor of degree $1$  since it has no zero in $\mathbb F_2$, so  there remains only to prove that $p(x)$ has no factor which is an irreducible polynomial $g(x)\in \mathbb F_2[x]$ of degree $2$.
But the only such  irreducible  polynomial is $g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved.