Exercise 5.6.8 (A double root of $p(x)\in \mathbb{Q}[x]$, where $\deg(p) \leq 3$, is rational.)

Proof. If $r\in ℂ$ is a double root of $p(x)$, then ${(x-r)}^2\mid p(x)$ in $ℂ[x]$, thus there are complex numbers $e,f$ such that

By expanding the product, we obtain

Therefore $e=a$, so

then $f-2\mathit{ar}=b$.

If $a=0$, then $e=0$ and $f\ne 0$ (otherwise $e=f=0$, and (1) shows that $a=b=c=d=0$, which is false by hypothesis), and $p(x)=f{(x-r)}^2$, so

Then $f=b$, and $c=-2\mathit{rf}$, thus $r=-\frac{c}{2f}=-\frac{c}{2b}\in \mathbb{Q}$.

We may suppose now that $a\ne 0$. The change of variable $x=y-\frac{b}{3a}$ transforms $p(x)=a{x}^3+b{x}^2+\mathit{cx}+d$ into $a(y^3+\mathit{py}+q)$, where $p,q$ are rational numbers, so that

(see for instance Cox “Galois Theory” Ex. 1.1.1 for details).

Put $s=r+\frac{b}{3a}$. Note that $r$ is rational if and only if $s$ is rational. By equation (1),

This shows that the multiplicity of $s$ in $q$ is at least $2$. Therefore

which gives

The elimination of $s$ between these two equations gives

• If $p=0$, then $q=0$, thus $s=0\in \mathbb{Q}$ by (2).
• If $p\ne 0$, then $s=-\frac{3q}{2p}\in \mathbb{Q}$.

In both cases, $s$ is a rational number. Therefore $r=s-\frac{b}{3a}$ is a rational number.

If $r$ is a double root of $p(x)=a{x}^3+b{x}^2+\mathit{cx}+d$, where $a,b,c,d$ are rational numbers, not all $0$, then $r$ is rational. □