Factoring Polynomials

Collin D collin.day.0 at gmail.com
Thu Dec 18 21:27:35 EST 2008


On Dec 18, 6:23 pm, "Russ P." <Russ.Paie... at gmail.com> wrote:
> On Dec 18, 6:17 pm, Collin D <collin.da... at gmail.com> wrote:
>
>
>
>
>
> > On Dec 18, 6:12 pm, Collin D <collin.da... at gmail.com> wrote:
>
> > > On Dec 18, 11:37 am, collin.da... at gmail.com wrote:
>
> > > > I am trying to write a simple application to factor polynomials. I
> > > > wrote (simple) raw_input lines to collect the a, b, and c values from
> > > > the user, but I dont know how to implement the quadratic equation
>
> > > > x = (-b +or- (b^2 - 4ac)^1/2) / 2a
>
> > > > into python. Any ideas?
>
> > > I completed the code:
>
> > > #import
> > > from math import sqrt
>
> > > # collect data
> > > a = float(raw_input('Type a value: '))
> > > b = float(raw_input('Type b value: '))
> > > c = float(raw_input('Type c value: '))
>
> > > # create solver
> > > def solver(a,b,c):
> > >     if b**2 - 4*a*c < 0:
> > >         return 'No real solution.'
> > >     else:
> > >         sol1 = -1 * b + (sqrt(b**2 - 4*a*c)) / 2*a
> > >         sol2 = -1 * b - (sqrt(b**2 - 4*a*c)) / 2*a
> > >         return (sol1, sol2)
>
> > > # execute
> > > print solver(a,b,c)
>
> > > Thanks to everyone who helped...
> > > This really expanded my knowledge on some of the mathematical
> > > functions in Python.
>
> > UPDATE:
> > '
>
> > #import
> > from math import sqrt
>
> > # collect data
> > a = float(raw_input('Type a value: '))
> > b = float(raw_input('Type b value: '))
> > c = float(raw_input('Type c value: '))
>
> > # create solver
> > def solver(a,b,c):
> >     if b**2 - 4*a*c < 0:
> >         return 'No real solution.'
> >     else:
> >         sol1 = (-1 * b + (sqrt(b**2 - 4*a*c))) / 2*a
> >         sol2 = (-1 * b - (sqrt(b**2 - 4*a*c))) / 2*a
> >         return (sol1, sol2)
>
> > # execute
> > print solver(a,b,c)
>
> You need to put your denominator, 2*a, in parens. The way it stands,
> you are dividing by 2, then multiplying by a. That's not what you
> want.
>
> Also, for better style, I suggest you compute the discriminanat once
> and store it for reuse rather than repeating the expression three
> times.- Hide quoted text -
>
> - Show quoted text -

I see what you mean on the denominator and discriminant. Ill do that.



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