Hello (about factoring into explicit algebraic numbers)

Your problem cannot be solved in general form mathematically,
and so you should not expect that some program would solve it.

You cannot express the roots of arbitrary polynomials in
terms of radicals, according to Galois. (For degree 5 or higher
there exist polynomials that cannot be solved in this form.)

You can find approximate linear factors by finding complex
roots numerically.

The "factor" commands do the best possible "over the rational field".
Which means that some polynomials are irreducible.

You can find some solutions in terms of radicals, but that is
only when the polynomial is of low enough degree, or you are
lucky.

Regards
RJF




Troumad wrote:

> Robert Jerrard a écrit :
>
>> On Thu, 2005-03-11 at 16:57 +0100, Troumad wrote:
>>  
>>
>>> I'm french and I don't speak english, but i try...
>>> I will like to be able to factorize fractions like x²-2, but that 
>>> does not go because the root is not whole, it's sqrt(2). It's good 
>>> with x²-4 because the root is 2 !
>>>   
>>
>>
>> Hi Troumad, perhaps some others can offer further information but I 
>> believe that factor is specifically working in the integers. The 
>> command describe(factor); shows that it is "over the field of 
>> integers" so you will not get it to show (x-sqrt(2))*(x+sqrt(2)). On 
>> the other hand you could do solve(x^2-2) which will give you plus and 
>> minus sqrt(2) as roots of the polynomial.
>> Hope this helps, Bob.
>>  
>>
> Yes... But I seek another method to factorize directly.  I have the same
> problem with the rational fractions.
> Lukic Milan has to give a good method, but it's not simple!
>
> I seek a simple software GPL to advise with my students to solve the
> problems that one gives them.
>