= Solution
For a depressed quartic
$$
f(x)=x^4+px^2+qx+r_0
$$
with roots summing to zero, set
$$
\beta_1=\alpha_1\alpha_2+\alpha_3\alpha_4,
\quad
\beta_2=\alpha_1\alpha_3+\alpha_2\alpha_4,
\quad
\beta_3=\alpha_1\alpha_4+\alpha_2\alpha_3.
$$
The <cubic resolvent of a quartic> is
$$
g(y)=\prod_{i=1}^3(y-\beta_i)
=y^3-py^2-4r_0y+(4pr_0-q^2).
$$
The symmetric group $S_4$ acts on the three partitions of four roots into two unordered pairs. The resulting homomorphism $S_4\to S_3$ has kernel the <Klein four-group> $V_4$. If $\Delta(f)$ is a square and the characteristic is not two, the <discriminant criterion for an alternating Galois group> gives $G=\operatorname{Gal}(f)\leq A_4$, and its image on the three resolvent roots lies in $A_4/V_4\cong C_3$.
For an irreducible quartic, $G$ is transitive on four roots, and its only possibilities inside $A_4$ are $V_4$ and $A_4$. The resolvent is irreducible exactly when the image acts transitively on its three roots, equivalently when that image is $C_3$. Thus
$$
g\text{ irreducible}\iff G=A_4,
\qquad
g\text{ reducible}\iff G=V_4.
$$
The printed statement omits the needed irreducibility hypothesis on $f$. Literally it is false: $f=x(x^3-3x-1)$ has square discriminant and Galois group $C_3$, and its resolvent is irreducible. Without quartic irreducibility, a reducible resolvent only implies $G\leq V_4$, allowing the trivial group, $C_2$, or $V_4$.
For the specified polynomial $x^4+8x+12$,
$$
\Delta=256\cdot12^3-27\cdot8^4=331776=576^2.
$$
Its resolvent is $g(y)=y^3-48y-64$. Modulo $5$ this is $y^3+2y+1$, which has no root in $\mathbb F_5$ and is therefore an <irreducible polynomial>[irreducible] cubic. The quartic is given to be irreducible, so the preceding criterion yields
$$
\boxed{\operatorname{Gal}(x^4+8x+12)=A_4}.
$$
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