The only two methods I know are
Eisenstein's method
Irreducibility modulo $n$
Now, I am asked the following question
Show whether or not $p(x)=x^5-5x^4+10x^3-7x^2+8x-4$ is irreducible over $\mathbb{Q}$ or not.
Eisenstein didn't help. No such $p$ prime can be found to satisfy the criterion(by the way, once I can show irreducibility over $\mathbb{Z}$, then Gauss' lemma tells me it's irreducible over $\mathbb{Q}$).
So method 2? Well I tried considering $p(x)$ over mod $5$ and mod $2$ which didn't help; they were both reducible.
Out of ideas and I looked at the solution, which said
$p(s+1)=s^5+3s^2+9s+3$ is irreducible by Eisenstein's criterion with $p=3$.
I'm just baffled, $p(s+1)$? $s+1$? What is this reasoning here? What's $s$? Is this some...integer?
Just, okay, someone tell me what it's doing, and why $p(s+1)$ being irreducible proves $p(x)$ being irreducible. Does it work for, I don't know, $p(s+2)$? $s+3$?
What is this method?
First observe the beginning of the polynomial is the beginning of the expansion of $$(x-1)^5=\color{red}{x^5-5x^4+10x^3}-10x^2+5x-1,$$ so we rewrite $p(x)$ as $$(x-1)^5+3(x^2+x-1).$$ Now set $s=x-1$, and write everything with $s$: $$p(x)=p(s+1)=s^5+3(s^2+3s+1).$$ Eisenstein's criterion says $p(s+1)$ is irreducible, hence $p(x)$ is, since $x\mapsto x-1$ defines an automorphism of $\mathbf Q[x]$.