I have an exercise from my course notes that states:
Find how many irreducible factors has $f(x) = x^{26}-1$ over $\mathbb{F}_3$ and their degrees. (don't factorize it)
I see immediately that the $1$ is a root of $f$. So I have $f(x)=(x-1) g(x)$ where $g$ has degree $25$ with no root in $\mathbb{F}_3$. But I don't know really how to move.
The distinct-degree factorization theorem tells us that $X^{27}-X$ factors mod $3$ into the product of all monic irreducible polynomials in $\mathbb{F}_3[X]$ whose degree divides $3$ (since $27 = 3^3$). So $X^{26}-1$ is the product of all monic irreducible polynomials in $\mathbb{F}_3[X]$ of degree $1$ and $3$ except $X$.