Factoring Polynomials

[Collin D]

On Dec 18, 1:09 pm, Mark Dickinson <dicki... at gmail.com> wrote:
> On Dec 18, 8:47 pm, Scott David Daniels <Scott.Dani... at Acm.Org> wrote:
>
> >      else: # a single result (discriminant is zero)
> >          return (-b / (2 * a),)
>
> Maybe make that (-b / (2. * a)) to avoid getting funny results
> when a and b are integers.  (Or do a from __future__ import
> division, or use Python 3.0, or ....)
>
> And to make the function more bullet-proof, you might want to
> do something like (untested):
>
>     from math import copysign
>
>     [rest of example as in Scott's post]
>
>     if discriminant: # two results
>         root1 = (-b - copysign(discriminant, b))/(2*a)
>         root2 = c/(a*root1)
>         return (root1, root2)
>
> to avoid numerical problems when b*b is much larger
> than abs(a*c). Compare with the results of the usual
> formula when a = c = 1, b = 10**9, for example.  But
> that still doesn't help you when the computation
> of the discriminant underflows or overflows...
>
> Isn't floating-point a wonderful thing!  :)
>
> Mark

Thanks for all your help! Its good to know how to do it w/ without
numpy.
And yes, floating point is the best thing since sliced bread. ^^

-CD