Let $E$ be the splitting field of $f(x)=x^4-x-2 $ over $\mathbb{Z_3}$ determine $[E:\mathbb{Z_3}]$ and factor $f(x)$ in linear factors in $E[x]$
It's clear that $2$ is a root of $f(x)$ in $\mathbb{Z_3}$, hence we can write $f(x)=(x-2)(x^3+2x^2+x+1)$ in $E[x]$, I'm unsure how I could go on from here since the cubic we got it's not easily factorable, it should have a real root and two complex conjugate solutions.
How can I determine $[E:\mathbb{Z_3}]$? and how could I factor further? The solution presented on the notes where I found the exercise states the following factorization: $f(x)=(x-2)(x-\alpha)(x-\alpha^3)(x-\alpha^9)$ where $\alpha$ is a root of $x^3+2x^2+x+1$
I assume that $\mathbb{Z}_3$ refers to the finite field with $3$ elements. First show that $g=x^3+2x^2+x+1$ is irreducible. Note that it is a degree $3$ polynomial, so it is sufficient to show that it has no roots (in $\mathbb{Z}_3$).
So you know there is a degree $3$ extension of $\mathbb{Z}_3$ in which $g$ has a root $\alpha$. Now show that this is already a splitting field of $g$ and therefore a splitting field of $f$ and determine the conjugates of $\alpha$ in order to factor $f$. Hint for this part: Extension of finite fields are Galois-extensions. What is the Galois group?