Let $f\in \mathbb{Q}[x]$ be an irreducible cubic, and let $p$ be a real root of $f$. Prove that the splitting field of $f$ over $\mathbb{Q}$ is $\mathbb{Q}(p, \sqrt{D})$, where $D$ is the discriminant of $f$. (Aluffi, Algebra: Chapter $0$, Chapter 7, Exercise 7.6)
I know that $p$ gives us the extension $\mathbb{Q}(p)$, but does that mean we can write $f$ and $(x \pm p$) (some irreducible quadratic)?
Let $at^3+bt^2+ct+d$ be a cubic equation. The change of variables $t=x-\frac{b}{3a}$ gives us $x^3+rx+s$ for some $r,s\in\mathbb{Q}$. Labelling the other two roots $p_1,p_2$, we have the following relations
$p+p_1+p_2=0$
$pp_1p_2=-s$
Therefore, $(x-p_1)(x-p_2)=x^2+px-\frac{s}{p}$, and also $\sqrt{D}=(p-p_1)(p-p_2)(p_1-p_2)=(2p^2-\frac{s}{p})(p_1-p_2)$.
This gives us that $p_1-p_2\in\mathbb{Q}(p,\sqrt{D})$, and also $p_1+p_2=-p\in\mathbb{Q}(p,\sqrt{D})$.
Therefore $\mathbb{Q}(p,p_1,p_2)\subset\mathbb{Q}(p,\sqrt{D})$ and the other inclusion is obvious.