Factor $x^5+x^2+1$ into irreducible polynomials in $Z[x]$ So here is my question: i would like to determine if whether or not the polynomial $x^5+x^2+1$ in $Z[x]$ is irreducible and if not then find the factors.
I tried a lot to find it. I really think it is already irreducible but I don't know how to prove it.
I tried Eisenstein's criteria which doesn't work here, I tried to check this criteria also after changing $x$ as $ax+b$ but everything I check was too complicated.
I tried for the modulus with prime numbers of $Z$, and I'm not sure of my following argument:
The polynomial $\mod 2$ stays $x^5+x^2+1$. 
But if $x$ is odd then $x^5$ and $x^2$ are odd too. Then the polynomial is given by $1+1+1=3=1 \mod 2$ which is irreducible (not sure if I have the right to say that).
And if $x$ is even, then $x^5$ and $x^2$ are even too. The polynomial is thus given by $0+0+1=1 \mod 2$ which is irreducible.
Hence the polynomial is irreducible $\mod 2$ and since it's a monic polynomial it implies it is irreducible in $Z[x]$
Is my argument correct?
If not if you have any clue please I would like to have some :)
Thank you and good night!
 A: Your argument that $\bar f=x^5+x^2+1$ is irreducible over $\Bbb F_2$ is not valid. As Thomas Andrews has noted, you’ve only shown that $\bar f$ has no linear factors. Fortunately, there is only one irreducible quadratic polynomial over $\Bbb F_2$, and that is $x^2+x+1$. But you easily check that when you divide this into $\bar f$, the quotient is $x^3+x^2$ and the remainder is $1$, so $\bar f$ has no linear nor quadratic factor, and therefore is irreducible.
As a result, your polynomial over $\Bbb Z$ is irreducible.
A: 
So here is my question: i would like to determine if whether or not the polynomial x5+x2+1 in Z[x] is irreducible and if not then find the factors.

Lubin already addressed the issue in your approach, let me show an alternative way to see it is actually irreducible using Cohn's irreducibility criterion: The $f(2)=37$ is a prime, and since the coefficients are non-negative and less than $2$, $f(x)$ is irreducible in $\mathbb{Z}[x]$. Another way to look at this criterion is that the coefficients of the polynomial written in base $2$ give a prime, i.e. $(100101)_2=37.$
