So this question will probably turn out to be way too trivial but I'd still like to get an answer because I'm not seeing it yet.
Given $f = X^{4} - 4X^{3} - 4X^{2} + 16X - 8$. I have (according to what the exercise asked me to do) shown that $(1/8)*X^{4}*f(2/X)$ is Eisenstein at 2 and therefore irreducible, but now I should see that this implies that f is also irreducible (in $\mathbb{Q}[X]$). You are seeing why? Thanks!
If $f$ were not irreducible, say $f(X) = g(X) h(X)$ for polynomials $g$ and $h$ over the rationals of degrees $d_g$ and $d_h$, both $\ge 1$, then $X^4 f(2/X)/8 = (X^{d_g} g(2/X)) (X^{d_h} h(2/X))/8$ where $X^{d_g} g(2/X)$ and $X^{d_h} h(2/X)$ would be polynomials over the rationals of degrees $d_g$ and $d_h$, so that would not be irreducible.