If $c$ is a complex root of a cubic $a(x)\in\mathbb{Q}[x]$, then $\mathbb{Q}(c)$ is the root field of $a(x)$ over $\mathbb{Q}$.

Question

This question originates from Pinter's Abstract Algebra, Chapter 31, Exercise C4.

Prove that if $c$ is a complex root of a cubic $a(x)\in\mathbb{Q}[x]$, then $\mathbb{Q}(c)$ is the root field of $a(x)$ over $\mathbb{Q}$.

Given $c$ is a complex root of $a(x)\in\mathbb{Q}[x]\subseteq\mathbb{R}[x]$, so is the conjugate $\bar{c}$. So $a(x)=(x-c)(x-\bar{c})b(x)$ over $\mathbb{Q}(c)$ for some polynomial $b(x)$ . Note $b(x)$ has degree 1, and so the root of $b(x)$ is in $\mathbb{Q}(c)$.

Is this a reasonable argument?

Update: Turns out this is a duplicate of Splitting Field of Cubic Polynomial Over the Rationals.