Factor $x^5+x^2+1$ into irreducible polynomials in $Z[x]$

Solution 1:

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.

Solution 2:

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. $(100101)_2=37.$