Re: -- reducibility in Q[x]

On Sat, 16 Feb 2008 18:19:48 EST, tommy1729 <tommy1729@xxxxxxxxx>
wrote:

i am amazed by your question quasi.

just take (x-a)(x-b)(x-c) =

and you can express the coefficients in the zero's.

this thus all leads to simple diophantines.

besides there are discriminants for all polynomials , and they dont just exist to solve the polynomials.

perhaps you want to look at the formula for solving deg 3 again...

Ok, I will.

But I want to do it over an arbitrary field.

I'll ask the question in a new thread.

quasi
.

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