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?

  • $\begingroup$ You have to solve a cubic to do it. In other words, it will not give you a "nicer" way. $\endgroup$ – André Nicolas Jun 8 '12 at 7:02
  • $\begingroup$ Is there any simple explanation as to why it works for the quartic but not for the cubic? I mean, for the quartic, one can reduce the problem to a cubic. Why is it not possible to reduce the cubic into a quadratic? $\endgroup$ – hedgepig Jun 8 '12 at 7:04
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    $\begingroup$ I do not know of one. There is a complicated Galois Theory reason. Sort of connected with the fact that $3$ is prime. $\endgroup$ – André Nicolas Jun 8 '12 at 7:08
  • $\begingroup$ In that case, do you mean that one must "solve the cubic" to solve the cubic? Or, does this only apply for the roots of the polynomial? $\endgroup$ – hedgepig Jun 8 '12 at 7:27
  • $\begingroup$ It applies equally to expressing as a product of linear term and quadratic. The linear term gives an immediate root, and after we know the coefficients of the quadratic, the other two roots are easy. $\endgroup$ – André Nicolas Jun 8 '12 at 12:25

I don't understand the question. Cardano's method does solve the cubic by replacing it with a quadratic. First you make a substitution to bring it to the form $x^3-px-q=0$; then you make another substitution that looks like turning it into an equation of degree 6, but on closer inspection you find it's a quadratic in the cube of the variable. So you have replaced the cubic with that quadratic.

  • $\begingroup$ The question has nothing to do with cardano's method. I was trying to factor the cubic into the a linear and quadratic factor. Andre Nicolas explained to me that I could not reduce the problem that way, however. $\endgroup$ – hedgepig Jun 8 '12 at 15:56
  • $\begingroup$ Your exact words: "Why is it not possible to reduce the cubic into a quadratic?" My answer: it is possible to reduce the cubic to a quadratic, and that's exactly what Cardano does. True, it doesn't factor the cubic into a linear and a quadratic, but that wasn't what you asked for in the comment I quoted. $\endgroup$ – Gerry Myerson Jun 9 '12 at 1:10
  • $\begingroup$ I do realize what exactly was wrong with how I phrased my words. I guess I should've said something closer to "Why is it not possible to factor a cubic into a linear and quadratic factor?" $\endgroup$ – hedgepig Jun 9 '12 at 7:09

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