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!
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.