 Solution

For a depressed quarticwith roots summing to zero, setThe cubic resolvent of a quartic is

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 $G=\mathrm{Gal}(f)\le A_4$, 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 $C_3$. ThusThe 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 $G\le V_4$, allowing the trivial group, $C_2$, or $V_4$.

For the specified polynomial $x^4+8x+12$,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 cubic. The quartic is given to be irreducible, so the preceding criterion yields