= Solution
Irreducibility modulo two implies irreducibility over $\mathbb Q$ by <Gauss lemma for polynomials>, so the <Galois group> embeds transitively in $S_4$. The <discriminant> is $229$, hence primes two and three are unramified. The factor-degree theorem says that square-free reduction with factor degrees $d_1,\ldots,d_r$ provides an element of the <Galois group> with cycle lengths $d_1,\ldots,d_r$. The reductions already found therefore give a 4-cycle and a 3-cycle.
The <group> order divides 24 and is divisible by four and three, so is 12 or 24. An order-12 subgroup of $S_4$ has index two and must be $A_4$: in its quotient every <transposition> has the same nontrivial image, since <transpositions> are conjugate and generate $S_4$, making the quotient map the sign. But a 4-cycle is odd. Therefore the order is 24 and
$$
\boxed{\operatorname{Gal}(f/\mathbb Q)=S_4.}
$$
The <discriminant> can also be obtained from $f'=x^2(4x+3)$ and the <resultant>, giving $256-27=229$.
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