# [Maxima] Factorization in extension fields

Stavros Macrakis macrakis at alum.mit.edu
Sun Jan 20 08:20:28 CST 2008

```? Factor will show you the documentation for 'factor'.

Try factor(p, q^2-2 ) for example.

-s

On 1/20/08, 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).  Its algorithm can be interpreted as finding
> factorizations if the field of rational numbers (by Gauss' lemma it suffices
> to find factorizations over the integers).  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?  One could do so manually for a
> fixed entension field by setting up a set of simultaneous equations to be
> solved, but I wonder if there is a quicker, more elegant approach, in which
> Maxima even discovers the right extension field.
> Thanks, George Leeman
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