Find a second root of $x^3+px+q$ given the first root

Question

This is a problem from Artin where given one root $a$, you have to find an equation for a second root in terms of $a$, $p$, $q$, and the square root of the discriminant $\delta$.

Here's what I have so far. The coefficients are the symmetric functions evaluated at the roots, so if the roots are $a$,$b$,$c$ then:

$s_1=a+b+c=0$

$s_2=ab+ac+bc=p$

$s_3=abc=-q$

And we can also use the square root of the discriminant,

$(a-b)(a-c)(b-c)=\delta=\sqrt{-p^3-27q^2}$

At this point it looks like its just doing a bunch of algebra to cancel $c$ and write $b$ in terms of $a,p,q,\delta$ but I've been at it for a while and don't seem to be having any luck.

Answer 1

We have $b+c = -a$ and $bc=-\frac{q}a$. This means $b$ and $c$ are roots of the quadratic $$y^2+ay-\frac{q}a = 0 \implies ay^2 + a^2y - q = 0$$

Answer 2

Another, perhaps less elegant approach is to perform division. Then,

$$\frac{x^3+px+q}{x-a}=x^2+ax+(a^2+p)+\frac{q+a(a^2+p)}{x^3+px+q}$$

Thus since $a$ is a root, we can eliminate $q$ as $q=-a(a^2+p)$.

Find the other roots, from the quadratic factor, in terms of $a$ and $p$.