For $p \neq 5$, I need to find the degrees of the factors of $f(x)=x^5-5$ in the cases where
a) $p \equiv 2$ or $3$ mod $5$ (show $f$ is the product of a linear factor and irreducible quartic over $\mathbb{F}_p$)
b) $p \equiv 1$ mod $5$ (show $f$ is either irreducible or splits completely)
So far, I have found the splitting field of $f$ over $\mathbb{Q}$ to be $\mathbb{Q}(\sqrt[5]{5},\omega)$ where $\omega$ is a primitive $5$th root of unity, and this splitting field has degree $20$ over $\mathbb{Q}$. This immediately shows that $f$ cannot have an irreducible cubic as a factor since $3$ does not divide $20$ yet we know $\text{Gal}(f/\mathbb{Q})$ contains an element of cycle type corresponding to the irreducible factorization of $f$ mod $\mathbb{F}_p$. However I am not sure how to generalize this to the cases where $p$ takes different values mod $5$ - my initial thought was to consider using Fermat's Little Theorem but working backwards from a general value of $p$ doesn't seem to be working. In addition, the criterion where you use the cycle types of the factors mod $p$ to find orders of elements in the Galois group also doesn't seem to help since we effectively already know what the group is.
There is really no Galois theory involved here. Consider the map $$\varphi:\ \Bbb{F}_p^{\times}\ \longrightarrow\ \Bbb{F}_p^{\times}:\ x\ \longmapsto\ x^5.$$ This is a group homomorphism of cyclic groups. If $p\not\equiv1\pmod{5}$ then it is injective, and so there is a unique $\alpha\in\Bbb{F}_p$ such that $\alpha^5=5$, meaning that $x^5-5$ has a unique linear factor. Then it remains to show that $x^5-5$ does not have a quadratic factor.
On the other hand, if $p\equiv1\pmod{5}$ then $|\ker\varphi|=5$, so either $x^5-5$ has no roots or it splits completely. Again it remains to show that $x^5-5$ does not have a quadratic factor.