 Solution

The quartic is irreducible by Eisenstein criterion at two, so its Galois group is transitive on its four roots. Its discriminant is $1616=16\cdot 101$, so five and seven are unramified primes for the polynomial. The squarefree factorizations areThe cubic modulo five has no root in ${\mathbb{F}}_5$, hence is irreducible. The quadratic modulo seven has discriminant five, a nonsquare there. The Frobenius cycle type theorem therefore supplies a three-cycle and a transposition in the group.

Transitivity implies that its order is divisible by four, and the three-cycle implies divisibility by three. Its order divides $24$, so it is $12$ or $24$. The only subgroup of index two in $S_4$ is $A_4$: an index-two subgroup is the kernel of a group homomorphism of a nontrivial homomorphism to $C_2$, and all transpositions are conjugate and generate $S_4$, forcing this homomorphism to be the sign. The exhibited transposition excludes $A_4$. Hence