       Re: Fourth degree polynomial

• To: mathgroup at smc.vnet.net
• Subject: [mg29285] Re: [mg29251] Fourth degree polynomial
• From: Niarlatotep <niarlatotep at ifrance.com>
• Date: Mon, 11 Jun 2001 04:38:27 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```on 8/06/01 10:15, Stephane Redon at Stephane.Redon at inria.fr wrote:

> Hello everybody,
>
> I've got a fourth degree polynomial which I would like to break into two
> second order polynomials. Unfortunately, the Factor function doesn't manage
> to do it, probably because it attempts to find all the roots of my
> polynomial. Is there a way to do this WITHOUT finding the roots ?
>
> Stephane
>
>
>
>

It's a marvelous problem...

I used an application to show to my son he could beat together his old Texas
Instrument TI-92 and Mathematica ! Since he was 14th he was very pleased.
Take x^5-1 and break it. He could do it by hand, the machine could
verify but not get it !

Let's return to the general problem, giving a fourth degree polynomial you
have three manners to break it into two second order polynomials.
Here is an simple example :
24 - 50*x + 35*x^2 - 10*x^3 + x^4 gives
(12 - 7*x + x^2)*(2 - 3*x + x^2) and
(8 - 6*x + x^2)*(3 - 4*x + x^2) and finally
(4 - 5*x + x^2)*(6 - 5*x + x^2).

The theory says a fourth degree polynomial has four roots, so it breaks into
(x-a)*(x-b)*(x-c)*(x-d) that gives three manners to break it into two second
order polynomials.

There is a way to do it without findings the roots but it's awful ! Finding
the roots is better.
Solve a fourth degree equation, the solution is very simple comparing to the
one solving
x^4 + a * x^3 + b * x^2 + c * x + d == (x^2 + l * x + m)(x^2 + p * q + r)
that's incredible !

Richard

```

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