Root field of $x^3+x^2+x+2$ over $\mathbb{Z}_3$

Question

Find the root field of $p(x)=x^3+x^2+x+2$ over $\mathbb{Z}_3$.

$p(x)$ is irreducible in $\mathbb{Z}_3$ by direct substitution of $x=0,1,2$.

Suppose $u$ is a root of $p(x)$. Then $[\mathbb{Z}_3(u):\mathbb{Z}_3]=3$ since the minimum polynomial has degree $3$.

Now, $p(x)=(x-u)q(x)$, where $q(x)=x^2+(u+1)x+(u^2+u+1)$.

Now I want to show that $q(x)$ is irreducible in $\mathbb{Z}_3(u)[x]$. I need to do direct substitution of the $27$ elements in $\mathbb{Z}_3(u)$. Is there an easier way?

Answer 1

Hint $\ q(x)\,$ has discriminant $ = u = (u^2-u)^2$