       Re: Factoring a polynomial

• To: mathgroup at smc.vnet.net
• Subject: [mg37111] Re: [mg37102] Factoring a polynomial
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Thu, 10 Oct 2002 03:20:42 -0400 (EDT)
• References: <200210090926.FAA17199@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Carlos Felippa wrote:
>
> 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?

The endless sci.math.symbolic thread strikes MathGroup!

Not exactly possible with no prior knowledge. Factor must work with a
given field, and the default is the rationals. You might direct it, say
by using the discriminant of the polynomial (as pointed out by Peter
Montgomery and Stephen Forrest on sci.math.symbolic. One may do this in
Mathematica as:

p1 = x^6+9/14*x^5+9/28*x^4+3/35*x^3+9/700*x^2+9/8750*x+3/87500;

I cribbed code for Discriminant right from www.mathworld.com:

Discriminant[p_?PolynomialQ,x_] := With[{n = Exponent[p,x]},

Cancel[((-1)^(n(n-1)/2)Resultant[p,D[p,x],x])/Coefficient[p,x,n]^(2n-1)]]

In:= InputForm[Factor[p1, Extension->Sqrt[Discriminant[p1,x]]]]

Out//InputForm=
((-15*I + Sqrt - (225*I)*x + 15*Sqrt*x - (1125*I)*x^2 +
75*Sqrt*x^2 - (3500*I)*x^3)*(15*I + Sqrt + (225*I)*x +
15*Sqrt*x + (1125*I)*x^2 + 75*Sqrt*x^2 +
(3500*I)*x^3))/12250000

Daniel Lichtblau
Wolfram Research

```

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