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MathGroup Archive 2002

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Re: Factoring question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35482] Re: Factoring question
  • From: Selwyn Hollis <slhollis at earthlink.net>
  • Date: Tue, 16 Jul 2002 04:49:39 -0400 (EDT)
  • References: <agbfhl$je9$1@smc.vnet.net> <200207130748.DAA08575@smc.vnet.net> <agrjfj$g9n$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Here's how this problem is suppose to play out:

You want to factor x^7 + x^5 + x^4 + -x^3 + x + 1. The rational root 
theorem tells you that the only possible rational roots are -1 and 1. 
You check those and find that -1 works. Therefore x+1 is a factor. Now 
you divide to find the other factor, which turns out to be 
1-x^3+2x^4-x^5+x^6. Since neither -1 nor 1 is a root of this factor, 
you're DONE---in the sense that you can't factor any more over the 
rational numbers. To go further, use NSolve to find the other six roots. 
None of them are real; so the previous factorization is complete over 
the irrationals as well.


Selwyn Hollis
slhollis at mac.com
http://www.math.armstrong.edu/faculty/hollis


Ken Levasseur wrote:
> 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: [mg35482]   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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