Re: Re: factoring quartic over radicals

*To*: mathgroup at smc.vnet.net*Subject*: [mg37072] Re: [mg37064] Re: factoring quartic over radicals*From*: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>*Date*: Tue, 8 Oct 2002 07:17:32 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Actually, including 1/GoldenRatio in the extension leads to an unnecessarily complicated formula. In this case there is no real need to so, since by definition In[30]:= Unevaluated[1/GoldenRatio==GoldenRatio-1]//FullSimplify Out[30]= True If one really insists on having the answer in the form proposed in Steve's original posting one can simply do: (Collect[#1, x] & ) /@ Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio}] /. -1 + GoldenRatio -> 1/GoldenRatio (-(-1 + x/GoldenRatio - x^2))*(1 + GoldenRatio*x + x^2) On Monday, October 7, 2002, at 06:26 PM, Jens-Peer Kuska wrote: > Hi, > > In[]:=Factor[x^4 + x^3 + x^2 + x + 1, Extension -> {GoldenRatio, > 1/GoldenRatio}] > Out[]=-((-3 - 2*x + Sqrt[5]*x + GoldenRatio*x - 3*x^2)* > (3 + x + Sqrt[5]*x + GoldenRatio*x + 3*x^2))/9 > > Regards > Jens > > Steve Earth wrote: >> >> Greetings MathGroup, >> >> My name is Steve Earth, and I am a new subscriber to this list and >> also a >> new user of Mathematica; so please forgive this rather simple >> question... >> >> I would like to enter the quartic x^4 + x^3 + x^2 + x + 1 into >> Mathematica >> and have it be able to tell me that it factors into >> >> (x^2 + GoldenRatio x + 1) ( x^2 - 1/GoldenRatio x + 1) >> >> What instructions do I need to execute to achieve this output? >> >> -Steve Earth >> Harker School >> http://www.harker.org/ > > > Andrzej Kozlowski Yokohama, Japan http://www.mimuw.edu.pl/~akoz/ http://platon.c.u-tokyo.ac.jp/andrzej/