Let $f(x)=x^3+1\in\mathbb F_5[x]$. (a) Write $f$ as a product of irreducibles and construct a splitting field $\mathbb{K}\subseteq L$ for $f$ over $\mathbb{F}_5$. (b) Determine the extension degree $[L:\mathbb{F}_5]$.
My thoughts are that $x^3+1=(x+1)(x^2-x+1)$ is such a product of irreducibles; the only way the quadratic factor would be reducible if it had a root in $\mathbb{F}_5$, but it is clear that no such root exists. Hence this is such a product. To construct a splitting field for $f$ over $\mathbb{F}_5$ we need to 'add' the roots of the polynomial which are not already in $\mathbb{F}_5$.
One root, $-1$, is certainly in $\mathbb{F}_5$ ($-1\equiv 4$ in $\mathbb{F}_5$). The other two roots are the roots of the quadratic factor, which are $\frac{-1\pm\sqrt{-3}}{2}$, and these certainly are not in $\mathbb{F}_5$. If we were to have the elements $\{i,\sqrt{3}\}$, however, we would indeed be able to solve the equation in $L$. But how do we write this splitting field?
For (b), I suspect the answer is $4$, as the minimum polynomial of $i$ is $x^2+1$ and the m.p. of $\sqrt{3}$ is $x^2-3$, both of degree $2$.
Are these answers correct?
Your factorization is correct but $[L:\mathbb F_5]=2.$
$L=\mathbb F_5[\sqrt{-3}].$