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MathGroup Archive 2002

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RE: Re: Factoring question

  • To: mathgroup at
  • Subject: [mg35576] RE: [mg35549] Re: Factoring question
  • From: "DrBob" <majort at>
  • Date: Fri, 19 Jul 2002 06:09:09 -0400 (EDT)
  • Reply-to: <drbob at>
  • Sender: owner-wri-mathgroup at

Steven has already said Mathematica can factor it; it just doesn't show
him HOW.


-----Original Message-----
From: Robert G. Wilson v [mailto:rgwv at] 
To: mathgroup at
Subject: [mg35576] [mg35549] Re: Factoring question

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
>just doesn't show how.  So, any ideas?
>"Ken Levasseur" <Kenneth_Levasseur at> wrote in message
>news:agrjfj$g9n$1 at
>>I assume that the problem was  to solve x^7 + x^5 + x^4 + -x^3 + x +
>>so, one of the basic factoring theorems is that if a polynomial over
>>integers  like this one has a rational root r/s, then r must divide
>>constant term and s must divide the leading coefficient.   So in this
>>+/-1 are the only possible rational roots and so the (x+1) factor
would be
>>found this way.   I'm sure that Mathematica checks this almost
>>As for the remaining 6th degree factor, I'm not certain how
>>proceeds, but if you plot it, it clearly has no linear factors.
>>Ken Levasseur
>>Steven Hodgen wrote:
>>>"DrBob" <majort at> wrote in message
>>>news:agbfhl$je9$1 at
>>>>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
>>>displays the factored result with no intermediate factoring steps.
>>>Thanks for the suggestion though.
>>>>-----Original Message-----
>>>>From: Steven Hodgen [mailto:shodgen at]
To: mathgroup at
>>>>Subject: [mg35576] [mg35549]   Factoring question
>>>>I just purchased Mathematica 4.1.  I'm taking precalculus and wanted
>>>>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
>>>>shows each step it used to get the the final result?

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