Show that \(\operatorname{Gal}(x^3+4x+2 / \mathbb{Q}) \cong S_3\).
Solution
Let \[ f(x)=x^3+4x+2. \] By Eisenstein’s criterion at \(2\), \(f\) is irreducible over \(\QQ\). Therefore its Galois group \(G\) acts transitively on the three roots, so \[ G\cong A_3\quad\text{or}\quad S_3. \]
For a depressed cubic \(x^3+px+q\), the discriminant is \[ \Delta=-4p^3-27q^2. \] Here \[ \Delta=-4\cdot4^3-27\cdot2^2=-364. \] This is not a square in \(\QQ\). For an irreducible cubic over a field of characteristic different from \(2\), the Galois group lies in \(A_3\) exactly when the discriminant is a square. Hence \(G\not\subseteq A_3\).
Therefore \[ \boxed{\Gal(f/\QQ)\cong S_3}. \]