Irreducibility of a polynomial over a field I'm trying to show that the polynomial $f(x) = \frac{x^5}{32}-3x-2$ is irreducible over $\mathbb Q$.
Obviously $f$ doesn't have a root over $\mathbb Q$ so I tried to use Gauss lemma for $32f$ and that led to nothing, also trying to use $x+a$ for several integers $a$ instead of $x$ in $f$ wasn't useful.
It seems like it would be the best to show that $f(2x)$ is irreducible (which is easily obtained by Eisenstein criteria), but it seems like subtituting $x$ with $2x$ will hurt the irreducibility (for instance consider $x\in \mathbb Z [x]$ is irreducible but $2x = (2)(x)$).
So what else can I try?

If $f(x)$ is reducible - say $f(x) = g(x)h(x)$, then $f(2x) = g(2x)h(2x)$ is reducible. So here it is fine to work with $f(2x)$.
Your counterexample doesn't work, as Gauss' Lemma - that irreducibility over $\mathbb Q$ is equivalent to irreducibility over $\mathbb Z$ - only works for monic polynomials. $2x$ is indeed irreducible over $\mathbb Q$.
And $f(2x) = x^5 - 6x - 2 $ is irreducible by Eisenstein's criterion.

In general, $f(x)$ is irreducible, if and only if  $f(ax+b)$ is irreducible by the same argument as above:$$f(x) = g(x)h(x) \implies f(ax + b) = g(ax + b)h(ax + b)\\f(ax + b) = g(ax + b)h(ax + b)\implies f(x) = f\left(\frac1a(ax+b)-\frac ba\right) = g(x)h(x)$$
This will, however, not work over $\mathbb Z$ if $a \ne \pm1,\ b\not\in\mathbb Z$, as the resulting polynomial will no longer be monic and/or in $\mathbb Z[X]$.
