Re: Re: Re: Factoring question

*To*: mathgroup at smc.vnet.net*Subject*: [mg35517] Re: [mg35485] Re: Re: Factoring question*From*: Andrzej Kozlowski <andrzej at yhc.att.ne.jp>*Date*: Wed, 17 Jul 2002 02:09:10 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Mathematica does not try to use any term groupings or re-arrangings that people usually try, because for computers much more direct methods are much faster and also guarantee getting the answer. In this case the expression is a quartic, so Mathematica can solve it in radicals using the formula of Cardano. The easiest way for a human who can't remember such complicated things is to notice that p and q are obviously roots of y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2==0. Thus (x-p)(x-q) must be a factor a factor. Once you know that the of the problem is trivial. Using Mathematica you can finish it in lots of ways, e.g. In[1]:= Cancel[((-p^2)*q^2 + (p^2*q + p*q^2)*y - (p + q)*y^3 + y^4)/ ((-p + y)*(-q + y))] Out[1]= (-p)*q + y^2 or In[2]:= PolynomialQuotient[y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2, (y - p)*(y - q), y] Out[21]= (-p)*q + y^2 or In[3]:= SolveAlways[y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2 == (a*y^2 + b*y + c)*((y - p)*(y - q)), y] Out[3]= {{c -> (-p)*q, a -> 1, b -> 0}, {c -> 0, a -> 1, b -> 0, p -> 0}, {c -> 0, a -> 1, b -> 0, q -> 0}} The last method is equivalent to what is sometimes called "comparing coefficients". On Tuesday, July 16, 2002, at 04:49 AM, Steven Hodgen wrote: > Actually, I never posted the actual problem, but here it is: > > y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2 > > This factors to: > > (y^2 - p*q)*(y - p)*(y - q) > > Which I cannot get. I've tried over an over again, and I'm the best > factorer in my class, my instructor can't seem to get this either, > hence my > hope that Mathematica could do it, and it certainly can no problem, but > it > just doesn't show how. So, any ideas? > > --Steven > > "Ken Levasseur" <Kenneth_Levasseur at uml.edu> wrote in message > news:agrjfj$g9n$1 at smc.vnet.net... >> Steve: >> >> I assume that the problem was to solve x^7 + x^5 + x^4 + -x^3 + x + >> 1=0. > If >> so, one of the basic factoring theorems is that if a polynomial over >> the >> integers like this one has a rational root r/s, then r must divide the >> constant term and s must divide the leading coefficient. So in this > problem, >> +/-1 are the only possible rational roots and so the (x+1) factor >> would be >> found this way. I'm sure that Mathematica checks this almost > immediately. >> As for the remaining 6th degree factor, I'm not certain how Mathematica >> proceeds, but if you plot it, it clearly has no linear factors. >> >> Ken Levasseur >> >> >> Steven Hodgen wrote: >> >>> "DrBob" <majort at cox-internet.com> wrote in message >>> news:agbfhl$je9$1 at smc.vnet.net... >>>> Factor[x^7 + x^5 + x^4 + -x^3 + x + 1] // Trace >>> >>> This doesn't do it. It only traces the initial evaluation, and then > simply >>> displays the factored result with no intermediate factoring steps. >>> >>> Thanks for the suggestion though. >>> >>>> >>>> Bobby >>>> >>>> -----Original Message----- >>>> From: Steven Hodgen [mailto:shodgen at mindspring.com] To: mathgroup at smc.vnet.net >>>> Subject: [mg35517] [mg35485] Factoring question >>>> >>>> Hello, >>>> >>>> I just purchased Mathematica 4.1. I'm taking precalculus and wanted > to >>> try >>>> a tough factoring problem, since the teacher couldn't do it either. >>>> Mathematica get's the correct answer, but I'm interrested in seeing > how it >>>> got there. Is there a way to turn on some sort of trace feature >>>> where > it >>>> shows each step it used to get the the final result? >>>> >>>> Thanks! >>>> >>>> --Steven >>>> >>>> >>>> >>>> >>>> >>>> >>>> >> >> > > > > > Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/

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