After learning Cardano's solution of the cubic, I decided to look at Ferrari's solution of the quartic (both articles on Wikipedia). At the end of the article on the quartic function was an an alternate method to solve the cubic by factoring into two quadratic terms - something like factoring the monic quartic $x^4+ax^3+bx^2+cx+d=0$ into $(x^2+px+q)(x^2+rx+s).$ They used Vieta's formula (correct me if I am wrong) in order to acquire the "resolvent" cubic (I assume that is a cubic required to be solved for algebraic solutions of the quartic) - what they essentially did was expanded the two quadratic factors, and set them equal to their corresponding coefficients.
I tried to do this with the monic cubic, first attempting to use three linear factors, then attempting to use one linear and one quadratic factor. Below I have divided by a for the first equation and expanded the second equation, setting coefficients equal.
$ax^3+bx^2+cx+d=0$ $(x+p)(x^2+qx+r)\;\longrightarrow \; p+q=0, \; r+pq=\frac{c}{a}, \; pr=\frac{d}{a}$ Note that $p+q=0$ when $b=0$, which can be the case of all cubics after application of the Tschirnhaus transformation.
However, I found that I could not solve this system - only piece it back together into a cubic. If there is a solution, how does one apply it?