Re: Factoring question

*To*: mathgroup at smc.vnet.net*Subject*: [mg35549] Re: Factoring question*From*: "Robert G. Wilson v" <rgwv at kspaint.com>*Date*: Thu, 18 Jul 2002 03:07:08 -0400 (EDT)*References*: <agbfhl$je9$1@smc.vnet.net> <200207130748.DAA08575@smc.vnet.net> <agrjfj$g9n$1@smc.vnet.net> <ah2457$r23$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Et al, I got Mathematica v4.1.0.0 to work Factor[ y^4 - (p + q)*y^3 + (p^2*q + p*q^2)*y - p^2*q^2] //Timing with output as required in 0.01 seconds. I also got the same result on my TI-89 in about a second. Good Luck, Bob. 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] >>>>Subject: [mg35549] 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

**References**:**Re: Factoring question***From:*"Steven Hodgen" <shodgen@mindspring.com>