Factoring Polynomials in Fields I always have problems to factorize polynomials that have no linear factors any more. For example ($x^5-1$) in $\mathbb{F}_{19}$. It's easy to find the root 1 and to split it. ($x^5-1$) = ($x-1$) * ($x^4$+$x^3$+$x^2$+x+1).
I think the last part must split into two irreducible polynomials with degree 2.
($x^2$+ ax+ b) ($x^2$+ Cx+ d). I expanded it and compared the coefficients to find
values for a,b,c,d. But it wasn't solvable. 
Is this approach correct or a there any other procedures or tricks to solve such a problem ? Thank you.

There is a trick to see that the polynomial is reducible. The multiplicative group $\mathbb{F}_{19^2}^*$ of the field of $19^2$ elements has $19^2 - 1 \equiv 0 \mod 5$ elements, so it contains a fifth root of unity. So the minimal polynomial of a fifth root of unity over $\mathbb{F}_{19}$, which divides $x^5 - 1$, has degree $2$. So your polynomial splits into quadratics.
I don't know a good way to find the factorization though, besides brute force. It turns out to be $x^5 - 1 = (x - 1)(x^2 + 5x + 1)(x^2 - 4x + 1)$.
