Let $f \in \mathbb{Q}[x]$ be an irreducible polynomial of degree 3. Suppose $f$ has one real root, we want to show that $$\text{Gal}(L/\mathbb{Q}) \cong S_3,$$ where $L$ is the splitting field of $f$.
Since $f(x)$ is irreducible, $f(x)$ has 3 distinct roots $r_1, r_2$ and $r_3$. If we consider the discriminant, we note that $$\Delta = \prod_{i < j} (r_j - r_i)^2.$$ Since the roots $r_2$ and $r_3$ are distinct, we have that $\sqrt{\Delta} \not \in \mathbb{Q}$.The splitting field of $f$ is given by $\mathbb{Q}(r_1, \sqrt{\Delta})$. By the tower law, we have that $$[\mathbb{Q}(r_1, \sqrt{\Delta}) : \mathbb{Q}] = [\mathbb{Q}(r_1, \sqrt{\Delta}) : \mathbb{Q}(\sqrt{\Delta})] \cdot [\mathbb{Q}(\sqrt{\Delta}) : \mathbb{Q}] = 3 \cdot 2 =6.$$ We therefore have that the Galois group of $L/\mathbb{Q}$ is isomorphic to $S_3$.
Is this is a legitimate proof?
Also, how can I generalise this result to prove that the Galois group of an irreducible polynomial of degree $p$ with $p-2$ roots is $S_p$?