= 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,
$$
f(X)\equiv X(X^3-X-2),
$$
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
$$
\boxed{\operatorname{Gal}(f/\mathbb Q)\cong A_4.}
$$
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