Re:Re: Re: Fourth degree polynomial

• To: mathgroup at smc.vnet.net
• Subject: [mg29306] Re:Re: [mg29285] Re: [mg29251] Fourth degree polynomial
• From: Niarlatotep <niarlatotep at ifrance.com>
• Date: Tue, 12 Jun 2001 04:18:24 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```>
> Your above example factors completely over the integers, and Factor will give
> you 4 linear factors. However, consider x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x
> +
> 1). Each of the quadratic polynomials on the right is irreducible, so this is
> the only factorization in this case, and the second method works better for
> this
> example.
>
> David
>

Dear David,

I gave a very simple example for every body to understand.

Theory is always right, you have three manners to break x^4 + x^2 + 1 :
you give the first (1 - x + x^2)*(1 + x + x^2)
this are the second and the third :
(-1 - I*Sqrt[3]*x + x^2)*(-1 + I*Sqrt[3]*x + x^2)
(1/2 - (I * Sqrt[3])/2 + x^2) * (1/2 + (I * Sqrt[3])/2 + x^2)

Have a good time,

Richard

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