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MathGroup Archive 2002

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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


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