Re: Re: Re: Re: Factoring question

*To*: mathgroup at smc.vnet.net*Subject*: [mg35581] Re: [mg35517] Re: [mg35485] Re: Re: Factoring question*From*: Andrzej Kozlowski <andrzej at bekkoame.ne.jp>*Date*: Fri, 19 Jul 2002 06:09:29 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Of course if If you really insist on solving this problem by grouping and factoring it is also quite simple: Consider the expression: Out[14]= (-p^2)*q^2 + p^2*q*y + p*q^2*y - p*y^3 - q*y^3 + y^4 group it as follows: (y^4 - p*y^3) + (p*q^2*y - p^2*q^2) + (p^2*q*y - q*y^3) factor out y-p: ((y - p))*y^3 + p*q^2*(y - p) - (y - p)*y*(p + y)*q hence (y-p)(y^3 + p*q^2 -y*(p + y)*q) re-group as (y-p)((y^3-y^2 q) + (p q^2-y pq)) factor out y-q: (y-p)((y-q)(y^2 -(y-p)pq) hence (y-p)(y-q)(y^2-pq) Andrzej Kozlowski On Wednesday, July 17, 2002, at 02:09 AM, Andrzej Kozlowski wrote: > 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: [mg35581] [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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**RE: Re: Factoring question**

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