Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 18H / iv / 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 $d_1,\dots ,d_r$ provides an element of the Galois group with cycle lengths $d_1,\dots ,d_r$. 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 andThe discriminant can also be obtained from $f^{\prime }=x^2(4x+3)$ and the resultant, giving $256-27=229$.