= Solution
A finite <solvable group> is a <group> admitting a finite chain of normal subgroups with abelian successive quotients; equivalently its derived series eventually becomes trivial. The chain
$$
S_4\triangleright A_4\triangleright V_4\triangleright1
$$
has quotients $C_2,C_3,V_4$, all abelian. Thus $S_4$ is solvable. Every <subgroup> of a <solvable group> is solvable, because intersecting the chain with the <subgroup> gives factors isomorphic to subgroups of the original abelian factors. A <Galois group of a polynomial> of degree at most $4$ embeds in $S_4$ by its action on the roots, so it is solvable even if the <polynomial> is reducible or has repeated roots.
For the substitution, the characteristic-zero <solvability by radicals> theorem says that a <polynomial> has solvable <Galois group> if and only if its <splitting field> is contained in a finite tower formed by adjoining successive roots of elements of the preceding <field>. First place all roots $\alpha$ of $f$ in such a tower. Adjoin a primitive $m$th <root of unity> and, for every nonzero $\alpha$, an $m$th root $\beta_\alpha$. These are radical adjunctions, and all roots of $x^m-\alpha$ are $\beta_\alpha\zeta_m^j$. Zero roots require no adjunction. The resulting radical tower therefore contains the <splitting field> of $f(x^m)$. The theorem implies
$$
\boxed{\operatorname{Gal}(f(x^m)/\mathbb Q)\text{ is solvable}.}
$$
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