Is $f(x)$ irreducible over $\mathbb{Z}_3$? I am looking for irreducible polynomials to construct finite fields. In this case I need a degree $4$ polynomial over $\mathbb{Z}_3$. Is $f(x)=x^4+x+2$ irreducible over $\mathbb{Z}_3$? I don't have much practice proving the irreducibility of a polynomial. So I am not sure what I have done. It's okay?
First, I have that $f(x)$ has no roots in $\mathbb{Z}_3$. If $f(x)$ was reducible, noting that there are no $x^3$ term in $f(x)$. So the factorization could be $(x^2+2x+c)(x^2+x+d)$, where $c,d \in \mathbb{Z}_3$ ($2 \equiv -1 \pmod 3$). So, $$x^4+x+2=x^4+(c+d+2)x^2+(c+2d)x+cd$$
Then $c+d+2=0$, $c+2d=1$ and $cd=2$. But of the first two equalities I have to $d=0$, wich is impossible since $cd\neq 0$. Thus $f(x)$ is irreducible over  $\mathbb{Z}_3$.

Assume that we have a factorization
$$x^4 + x + e= (x^2 + a x + b)(x^2 + c x + d)$$
Looking at the coefficient of $x^3$ we get $c=-a$. Now the coefficient of $x^2$ on RHS is $-a^2 + b + d$ so $d = a^2- b$. Now we get
$$(x^2 + a x + b)(x^2 - a x + (a^2 - b))= x^4 + (a^3 - 2 a b)x + a^2 b - b^2 $$
Now we get the system
$$a^3 - 2 a b = 1 \\
a^2 b - b^2 = e $$
Express $b$ from the last equation (assume $2$ invertible) and substitute in the second one. We get
$$a^6 - 4 e a^2 -1 = 0$$
Now in our case $e=2$, so the equation is $a^6 - 8 a^2 - 1 = 0$.  Moreover, over $\mathbb{F}_3$ we have $x^3 = x$ for all $x$, so $a^6 = a^2$. We get the equation $a^2 = -1$, not possible.
