 Solution

Modulo two the quartic has no root, and the only irreducible monic quadratic is $x^2+x+1$. Reducing the quartic modulo that quadratic gives remainder one. Hence the quartic is irreducible over ${\mathbb{F}}_2$. Its splitting field is ${\mathbb{F}}_{16}$, and $\boxed{{\mathrm{Gal}}_{{\mathbb{F}}_2}\cong C_4}$, generated by Frobenius.

Modulo three it factors as $(x-1)(x^3+x^2+x+2)$. The cubic has no root at zero, one or two and is irreducible, so its splitting field is ${\mathbb{F}}_{27}$ and $\boxed{{\mathrm{Gal}}_{{\mathbb{F}}_3}\cong C_3}$. Both reductions are square-free. Irreducibility modulo two makes the rational quartic irreducible, and its rational Galois group is transitive on four roots. The unramified factorization theorem supplies a four-cycle from the modulo-two reduction and a three-cycle from the modulo-three reduction. Its order is divisible by twelve and divides twenty-four. An order-twelve subgroup would have index two and equal $A_4$, which has no four-cycle. Thus $\boxed{{\mathrm{Gal}}_{\mathbb{Q}}\cong S_4}$.