As written in the title, I have to compute the splitting field of $$(x^3-x^2-x)(x^4-x^2+1)$$ over $\mathbb{F}_3$
I'd like to understand if my attempt is correst, or if I'm missing something. Here's what I did:
First of all, let's call $h(x)$ the first factor, and $g(x)$ the other one.
I focus on $h(x)$: the only root is $0$. Therefore, I can write $h(x)=(x-0) \cdot h_2(x)$ where $h_2(x)$ is a poly of degree $2$ with no roots in $\mathbb{F}_3$, hence it's irreducible in $\mathbb{F}_3$. Therefore, the splitting field of $h(x)$ over $\mathbb{F}_3$ is $\mathbb{F}_{3^2}=\mathbb{F}_{9}$
Now I focus on $g(x)$: it can be written eventually only as products of two irreducible quadratic polynomials with coefficients in $\mathbb{F}_3$, i.e.
$$x^4 + 2x^2 + 1 = (a x^2 + bx+c)(dx^2 + ex + f)$$ (I used that $-1 = 2 \text{mod}3$)
The conditions I get are
\begin{cases} ad = 1\\ ae+bd=0 \\ af + be + cd = 2 \\ bf + ce =0 \\ cf = 1 \end{cases}
and doing the computations I found $a=d=c=f=1, b=e=0$. Therefore, it can be written as product of two irred. quadratic polynomials. Therefore, we have that the splitting field of $g$ is $\mathbb{F}_{3^2}$
Now, in order to find the splitting field of the original polynomial, I note that the splitting fields of the factors are the same, so it is exactly $\mathbb{F}_9$