Re: Cubic eqn help

From: Oscar Lanzi III (ol3_at_webtv.net)
Date: 01/16/05 

Date: Sat, 15 Jan 2005 19:25:28 -0600

You have a real mouthful here in that "deduction" question.

The crux of the matter is that given the Cardano-Viete solution to the
equation, you can't express it in terms of real radicals. This is what
the whole casus irreducibilis thing is about. If real-radical solutions
exist, they have to be derived by a method other than the Cardano-Viete
formula -- meaning it's hit-or-miss because other methods, based on
conventional factoring techniques, are not guaranteed to work.

So you need to solve the cubic equation in two ways. One is the
Cardano-Viete formula, getting the roots in terms of trig and inverse
trig functions. Then you go back and solve the equaiton by the
elementary factoring approach, seeking a rational root. Of course you
know that the numerator in such a root has to divide the constant term,
the denominator has to divide the leading coefficient, and the root may
have either sign. If you find such a rational root r, divide out the
corresponding factor x-r and solve the remaining quadratic equation.

You now have two alternate sets of representations for the roots.
Equating roots between the two sets of representations (you have to be
sure, of course, that you pair them up correctly) should then give you
the desired result.

--OL