A quadratic or a cubic can be shown to be irreducible in $\mathbb{F}_p$ by showing that none of $0, \dots,p-1$ are roots (or more generally all the elements of $\mathbb{F}_{p^n}$). This does not work for quartics (and above) since it may factor into two irreducible quadratics (or higher), it only rules out the possibility of a linear and cubic (or one degree less) factorisation.
I'm new to doing stuff like this, and I've recently had to (for a larger problem) try to find whether or not a quartic is irreducible in $\mathbb{F}_5$. Specifically, my quartic was $X^{4}+2X^{3}+4X^{2}+3X+2$, but I'm asking for any general quartic. It has no roots in $\mathbb{F}_5$, so we have to check the products of quadratics.
In $\mathbb{F}_5$, there are $10$ such quadratics, which makes it fairly annoying to check, and in general is even larger for $\mathbb{F}_{p^n}$ (there are $\frac{1}{2}p^n(p^n-1)$) which seems very unpleasant to do for even some smaller prime powers. Of course we can rule out some possibilities by looking at the constant terms and requiring they multiply together to get the constant term of the quartic, reducing the possibilities significantly, but it's still a lot to check. Is there a better way to do this for quartics (or even higher polynomials)?