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MathGroup Archive 2002

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Re: Factoring a polynomial

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37110] Re: [mg37102] Factoring a polynomial
  • From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
  • Date: Thu, 10 Oct 2002 03:20:40 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On second thoughts:  any one who really wants to see what the exact 
answer is, can evaluate the following:


sols = Select[{a, b} /. SolveAlways[
      x^6 + (9/14)*x^5 + (9/28)*x^4 + (3/35)*x^3 +
        (9/700)*x^2 + (9/8750)*x + 3/87500 ==
       (x^2 + a*x + b)*(x^4 + c*x^3 + d*x^2 + e*x + f), x],
    FreeQ[N[#1], _Complex] & ]

Times @@ (Map[x^2 + {x, 1}.# &, sols])


N[%]


(0.010974992601737203 + 0.20255610310498245*x + x^2)*
   (0.020476912388332696 + 0.20476912388332683*x + x^2)*
   (0.15256133957420942 + 0.23553164887424147*x + x^2)

Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/


On Wednesday, October 9, 2002, at 08:56 PM, Andrzej Kozlowski wrote:

> Factoring without specifying the extension does not really make sense. 
> Of course Mathematica can easily factor yur polynomial into linear 
> factors over the complex numbers (with the help of Solve), but I 
> suspect you are really asking for is factoring over the reals. This is 
> harder and needs more human input. But anyway, Mathematica can do 
> this, or at least I have done it using Mathematica. In fact if you are 
> satisfied with a numerical answer Mathematica can do alone and in 
> seconds:
>
> In[1]:=
> Simplify[N[x^6 + (9/14)*x^5 + (9/28)*x^4 + (3/35)*x^3 + (9/700)*x^2 + 
> (9/8750)*x + 3/87500]]
>
> Out[1]=
> 1.*(0.010974992601737198 + 0.20255610310498295*x + 
> x^2)*(0.020476912388332692 +
>    0.2047691238833268*x + x^2)*(0.15256133957420948 + 
> 0.23553191586883315*x + x^2)
>
> But I have in fact been foolish enough to compute the exact answer 
> too. I do not propose to post it here for it's absolutely horrible 
> (expressed in terms of Root objects) and quite useless. However if you 
> really want to see it I can send it to you privately.
>
> Andrzej Kozlowski
> Yokohama, Japan
> http://www.mimuw.edu.pl/~akoz/
> http://platon.c.u-tokyo.ac.jp/andrzej/
>
>
>
> On Wednesday, October 9, 2002, at 06:26 PM, Carlos Felippa wrote:
>
>> From: carlos at colorado.edu (Carlos Felippa)
To: mathgroup at smc.vnet.net
>> Date: Wed Oct 9, 2002  6:26:02 PM Asia/Tokyo
>> To: mathgroup at smc.vnet.net
>> Subject: [mg37110] [mg37102] Factoring a polynomial
>>
>> Can Mathematica factor the polynomial
>>
>> p1=x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500;
>>
>> without a priori knowledge of the Extension field?
>>
>>
>>
>
>



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