Solution
= Solution
Irreducibility makes the quartic Galois group a transitive subgroup of $S_4$. Its square discriminant puts it inside $A_4$, leaving $V_4$ or $A_4$. The cubic resolvent is
$$
y^3-48y-64.
$$
The <rational root theorem> shows it is irreducible over $\mathbb Q$. Thus the Galois group acts transitively on the three pairings of four roots, excluding $V_4$. Consequently
$$
\boxed{\operatorname{Gal}(x^4+8x+12/\mathbb Q)\cong A_4}.
$$