TITLE: Proving that a polynomial is irreducible over a field
QUESTION [22 upvotes]: What's the general strategy to show that a particular polynomial is irreducible over a field?
For example, how can I show $x^4 - 10x^2 -19$ is irreducible over $\mathbb Q$?

REPLY [23 votes]: When a polynomial is integer valued, one may appeal to Gauss's lemma which states that if the coefficients of a non-constant polynomial $f$ are relatively prime and $f$ is irreducible in $\mathbb{Z}[X],$ then $f$ is irreducible in $\mathbb{Q}[X].$ 
This allows you to use the structure of $\mathbb{Z}[X]$ which is richer than that of $\mathbb{Q}[X]$ to prove irreducibility. For example for every prime $p$ there is a reduction map from $\mathbb{Z}[X]$ to $\mathbb{F}_p[X].$ If the reduction of $f$ is irreducible over $\mathbb{F}_p[X]$ then $f$ is irreducible over $\mathbb{Z}[X].$
Note that for your polynomial, $x^4 - 10x^2 - 19 \mod 3$ is irreducible in $\mathbb{F}_3[X].$ Hence, $x^4 - 10x^2 - 19$ is irreducible in $\mathbb{Z}[X].$

REPLY [6 votes]: Suppose that $p$ is a fourth degree polynomial which is not irreducible. Then $p$ contains a linear factor or a quadratic factor.
In your case you can easily check using the abc-formula that $p$ does not have a linear factor, i.e., it does not have a root lying in $\mathbf{Q}$.
Thus, now you have to think about why $p$ doesn't have a quadratic factor. You could do this with brute force, i.e., write out $p = (a+bx+x^2)(d+ex+x^2)$ and get to a contradiction. (This used to be $(a+bx+cx^2)(d+ex+fx^2)$, but as lhf points out it's clear that one may assume that $c=f=1$.)
Easier would be to  apply one of the methods mentioned by Ragib Zaman.
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