Show that for all positive integers $n,$ $$(x^2 + 2)^n + 5(x^{2n-1} + 10x^2 + 5)$$ is irreducible in $\mathbb{Q}[x].$
I think Schonemann's criterion finishes off the problem but I'm not sure how to show that $x^2 + 2 \pmod{5}$ doesn't divide $x^{2n-1} + 10x^2 + 5.$
Using the criterion you mention, the only thing left to show is that $x^2 + 2$ does not divide $x^{2n - 1} + 10x^2 + 5$ over the field $\Bbb F_5$. (This is just a fancy way of saying "modulo $5$".)
However, note that modulo $5$, we have $$x^{2n - 1} + 10x^2 + 5 \equiv x^{2n - 1}.$$
Since $\Bbb F_5[x]$ is a UFD, we can conclude directly that $x^2 + 2$ does not divide $x^{2n - 1}$ and this finishes the job.