REPOST: Re: Reducing a Cubic to a Quadratic

From: Sylvain Croussette (NO_SPAM_sylvaincroussette_at_yahoo.ca)
Date: 03/17/05 

Date: 17 Mar 2005 16:09:59 GMT

John Schutkeker <jschutkeker@sbcglobal.net.nospam> dixit:

>"Larry Hammick" <larryhammick@OMIT-MEtelus.net> wrote in
>news:SxuZd.36924$ZO2.31957@edtnps84:
>
>> "John Schutkeker" <jschutkeker@sbcglobal.net.nospam> wrote in message
>> news:Xns9619BE8744398lkajehoriuasldfjknak@151.164.30.44...
>
>>> Does anybody know if Cardano's Method for solving
>>> cubics can be used to factor a cubic into
>>> (x-a)*(x^2+bx+c)=0?
>
>> Well yes in the sense that it gives all three roots u,v,w, and
>> so the cubic is (x-u)(xx - (v+w)x + vw).
>> Maybe you're thinking of the case of a cubic with real
>> coefficients and only one real root? But alas Cardano's
>> formulas express the roots as functions of complex numbers,
>> whether or not there is just one real root.
>
>That is pretty much what I was thinking of.
>
>Am I correct in interpreting your answer as saying that u is derived by
>a different procedure than v & w, since u *must* always be real, while v
>& w can be either real or complex?
>
>Since complex roots are (probably) not physical there's a good chance
>that the system I'm studying will always have only one root. So all I
>really need is a way to find u alone, and a way to prove that the other
>two are complex.
>
>>From your equation, I can just stick vw and -(v+w) into the quadratic
>formula to solve my second problem. But I still need u to solve the
>first.
>
>Would you know if it's possible to simply look up expressions for u, v &
>w, without having to tear my hair out studying Cardano's method?

It's possible to compute the real root without using complex numbers,
see http://www.geocities.com/scroussette/Cardanoe.txt

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