How to prove $f(x) = 3 + 9x^2 + 3x^5$ is irreducible over $\Bbb Z[x]$?

Question

I know Eisenstein criterion won't work as the polynomial is not primitive. I don't know what other techniques I can use.

Answer 1

$g(x)=x^5+3x^2+1$ is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{F}_2$.
$g(x)$ has no linear factors in $\mathbb{F}_2[x]$ since it has no root in $\mathbb{F}_2$, and the only quadratic irreducible polynomial over $\mathbb{F}_2$ is $x^2+x+1$, that is not a divisor of $g(x)$, since in $\mathbb{F}_2$ $$ x^5+x^2+1 = x^2(x^3+1)+1 = x^2(x+1)(x^2+x+1)+1.$$