factor over complex numbers?



On Jan 27, 2008 4:01 PM, Edwin Woollett <woollett at charter.net> wrote:

> >> Is there any option I can set to get factor(expr1) -> (-%i + x)^5  (%i
> + x)^5
> >> instead of getting (1 + x^2)^5 ?
> >factor(expr1,x^2+1);
>
> I get factor(expr1,x^2+1);  --> 0
>

The meaning of factor(ppp,qqq) is "factor ppp over the polynomial ring where
qqq=0".

For example:

    factor (  x^2 - 2*p*x +2 , p^2-2 ) =>  (x - p)^2

So if you do factor(x^3+1,x^3+1) you will naturally get 0, since you've told
Maxima that x^3+1 = 0.

If you want to introduce a new quantity, you need to use a different
variable name, e.g.

     factor(x^4+1,q^2+1) => (x^2 - q) * (x^2 + q)

           -s