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.

Your suspicion is correct: $(x^5 - 1)$ does split into a linear factor and two quadratic ones. Your approach even works out!
$(x^2 + ax + b)(x^2 + bx + c) = x^4 + (a+c)x^3 + (b + ac + d)x^2 + (ad+bc)x + bd$
So we have
$bd = 1$, $a + c = 1$, $ad+bc = 1$, $b + ac + d = 1$.
Immediately we derive
$d = \frac{1}{b}$, $c = 1-a$.
We then substitute in $ad+bc = 1$ to obtain
$a = \frac{1-b}{\frac{1}{b}-b}$
This alerts us that we have to be careful to check $b \in \{1,18\}$ separately.
If $b$ happens to be $1$, then we get that $1 = b + ac + d = 1 + ac + 1 = 2+ac$, and so $1 + ac = 0$. As it happens, this equation and $c = 1-a$ are satisfied when $a$ is $5$, yielding $c$ as $-4$. This gives our two irreducibles.
