On Jan 20, 2008 8:26 AM, George Leeman <george.leeman at sbcglobal.net> wrote:
> If we let p=10*x^2+17*x+3, Maxima properly computes
> factor(p)=(2*x+3)*(5*x+1)....
> ...However, in the field Q(sqrt(2)), i.e. all numbers of the form
> r1+r2*sqrt(2), where r1 and r2 are
> rational, another factorization is possible, e.g.
> p=(-2+2*sqrt(2))*x-3+3*sqrt(2))*((5+5*sqrt(2))*x+1+sqrt(2)). Is it possible
> to get Maxima to find such factorizations?
Sorry, I didn't read your original email thoroughly enough.
The factorization you give,
((-2+2*sqrt(2))*x-3+3*sqrt(2))*((5+5*sqrt(2))*x+1+sqrt(2));
is simply r*(2*x+3)*s*(5*x+1), where r=sqrt(2)-1 and s=sqrt(2)+1; r*s=1.
This is not a distinct factorization, any more than (2*x+2)*(x-1) and
(x+2)*(2*x-2) are distinct factorizations of 2*x^2-2.
As for the general question, if you have a polynomial irreducible over the
integers, you can always express its roots in terms of one of its roots e.g
.:
factor( x^5-x-1, q^5-q-1 ) => (x-q)*(x^4+q*x^3+q^2*x^2+q^3*x+q^4-1)
or
factor(x^8+1,q^8+1) =>
(x-q)*(x+q)*(x-q^3)*(x+q^3)*(x-q^5)*(x+q^5)*(x-q^7)*(x+q^7)
-s