I know that a root $\alpha$ is constructible if [$Q(\alpha):Q$] = $2^n$ for some integer n. If $\Sigma$ is the splitting field of $f(x)$, $n \leq [\Sigma : Q] =Gal(f(x)) \leq n!$ so, in this case, $2^n \leq [\Sigma : Q] = Gal(f(x)) \leq 2^n!$
I want to somehow use this to show that $Gal(f(x))$ is solvable, but I'm unsure of where to go from here. Any hints or suggestions would be greatly appreciated. Thank you!