 Solution

Dedekind's factorization/Frobenius theorem says that for a prime not dividing the discriminant, the degrees of the distinct irreducible factors modulo $p$ give the cycle lengths of an element of the Galois group.

The polynomial is irreducible over $\mathbb{Q}$ by the rational-root test followed by a comparison of possible monic quadratic factors. Its discriminant is $547^2$, a square, so its transitive Galois group lies in $A_4$. Modulo five,and the cubic has no root in ${\mathbb{F}}_5$, giving a three-cycle. A transitive subgroup of $A_4$ containing a three-cycle cannot be the Klein four group; hence