| Right. Any vaguely rigorous derivation of the cubic formula requires dipping into complex numbers, because the discriminant of the cubic can be negative. You can just choose to ignore that case and focus only on nonnegative discriminants, but that's no fun, is it? :-) Although, that is how it was "classically" done. If you want to find the complex roots, but don't want to deal with them in the derivation, you can just take the real root and multiply it by (-1 ± i √3)/2. Knowing that all you have to do is this, of course involves some much heavier algebra than would have been available classically, but, at least it's easy for students to understand (if they have the proper group theory and/or Galois theory background), or, at least to understand a plausibility argument. After looking at Wikipedia, I found that Lagrange had an interesting method[0] I didn't know about, in which he considers the discrete Fourier transform of the roots rather than the roots themselves. That approach makes some sense, because the whole power of the Fourier transform is to turn gnarly multiplication problems into simple addition problems. I also like this method, because it's honest about needing to dip into the complex numbers to find all roots. Also, it's kind of a cool historical note, because he was trying to solve the general problem of finding all roots of a polynomial, and hoped this method might generalize. Edit: here's a lesson plan from Stanford that walks through all the necessary algebra in 5 class periods: https://web.stanford.edu/~aaronlan/assets/symmetries-and-pol... --- [0]: https://en.wikipedia.org/wiki/Cubic_equation#Lagrange's_meth... |