MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Factoring a polynomial

  • To: mathgroup at
  • Subject: [mg37106] Re: [mg37102] Factoring a polynomial
  • From: Andrzej Kozlowski <andrzej at>
  • Date: Thu, 10 Oct 2002 03:20:33 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

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 

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]]

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

On Wednesday, October 9, 2002, at 06:26 PM, Carlos Felippa wrote:

> From: carlos at (Carlos Felippa)
To: mathgroup at
> Date: Wed Oct 9, 2002  6:26:02 PM Asia/Tokyo
> To: mathgroup at
> Subject: [mg37106] [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?

  • Prev by Date: Re: Mathematica Link for Excel - Problems in starting the link
  • Next by Date: Re: Factoring a polynomial
  • Previous by thread: Re: Factoring a polynomial
  • Next by thread: Re: Factoring a polynomial