From: hbaker@netcom.com (Henry Baker)
Newsgroups: sci.math.research
Subject: Re: Quartic Reduction
Date: Mon, 7 Oct 1996 05:26:16 GMT

In article <53485q$mvl@earth.njcc.com>, nahay@pluto.njcc.com (John Nahay) wrote:

> This is a standard algebraic manipulation. But, none of my books has it.
> 
> How do I reduce a general quartic: ax^4+bx^3+cx^2+dx+e=0
> to a trinomial quartic fz^4+gz+h=0?
> Can one use just rational transformations, or are algebraic (roots) 
> necessary?
> Feel free to e-mail me, too.
> John

You can reduce your general quartic more easily to the form:

fz^4+gz^2+h=0

which then shows you how to solve the equation, since you now have a
quadratic in the 'variable' z^2.

Doing this can be done by a Moebius mapping x -> (Az+B)/(Cz+D), where
A,B,C,D are (in general) complex numbers.  However, computing A,B,C,D
requires solving a cubic polynomial.

So a rational transformation works, but computing the coefficient of
the transformation is what requires the additional root operations.

See:

ftp://ftp.netcom.com/pub/hb/hbaker/quaternion/mobius/Graber-quartic.dvi.gz

(also .ps.gz).
==============================================================================

From: mtrott@wolfram.com (Michael Trott)
Newsgroups: sci.math.research
Subject: Re: Quartic Reduction
Date: 7 Oct 1996 12:52:49 GMT

In article <53485q$mvl@earth.njcc.com> nahay@pluto.njcc.com (John Nahay)  
writes:
> This is a standard algebraic manipulation. But, none of my books has it.
> 
> How do I reduce a general quartic: ax^4+bx^3+cx^2+dx+e=0
> to a trinomial quartic fz^4+gz+h=0?
> Can one use just rational transformations, or are algebraic (roots) 
> necessary?
> Feel free to e-mail me, too.
> John

Use a Tschirnhaus transformation 
(W. E. Tschirnhaus: Methodus auferendi omnes terminos intermedios ex data  
equatione, Acta Eruditorium 2, 204 (1683)). You need a radical for this.
--
Michael Trott
Wolfram Research, Inc.