= Solution
The <discriminant> is a <symmetric polynomial> in the four roots, so the <Fundamental theorem of symmetric polynomials> expresses it polynomially in the <elementary symmetric polynomial>. For this depressed quartic they are $e_1=e_2=0$, $e_3=-r$, $e_4=s$. Scaling each root by $t$ multiplies the <discriminant> by $t^{12}$, while $r,s$ scale by $t^3,t^4$. Thus only weighted-degree-twelve monomials can appear, giving $\Delta_f=Ar^4+Bs^3$.
For $r=0$, use $f'=4X^3$ and the monic <discriminant> identity $\Delta_f=(-1)^{n(n-1)/2}\prod_if'(\alpha_i)$. At $n=4$ this gives $4^4(\prod_i\alpha_i)^3=256s^3$, so $B=256$. For $s=0$, the roots are zero together with those of $g=X^3+r$. The cubic's <discriminant> is $-\prod g'(\beta_i)=-27r^2$. The extra squared differences from zero multiply this by $(\prod\beta_i)^2=r^2$, giving $-27r^4$. Hence
$$
\boxed{\Delta_f=256s^3-27r^4.}
$$
These computations are <polynomial> identities, so they extend to parameter values with repeated roots as well.
Back to article page
