MathGroup Archive 1995

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: [Q] Factor


f[x_] = (a + b x + c x^2) (2x - 2x^2);

Treat the polynomial in x as if it were a polynomial in a, b, or c. 
Generate the coefficients and keep only the first order term:

Last[CoefficientList[f[x], #]]& /@ {a, b, c}

          2     2      3     3      4
{2 x - 2 x , 2 x  - 2 x , 2 x  - 2 x }

Factoring does not produce the exact form requested:

Factor[Last[CoefficientList[f[x], #]]]& /@ {a, b, c}

                         2             3
{2 (1 - x) x, 2 (1 - x) x , 2 (1 - x) x }

Dividing by the desired factor produces the required "coefficients":

Cancel[(Last[CoefficientList[f[x], #]]& /@ 
	{a, b, c})/(2x - 2x^2)]

        2
{1, x, x }

Reintroducing the factor provides the required form of the factors
for a, b, and c:

Cancel[(Last[CoefficientList[f[x], #]]& /@ 
	{a, b, c})/(2x - 2x^2)] (2x - 2x^2)

          2              2    2           2
{2 x - 2 x , x (2 x - 2 x ), x  (2 x - 2 x )}


Bob Hanlon

>  From: blast at bendix.swb.de (Bernd Last)
>  Newsgroups: comp.soft-sys.math.mathematica
>  Subject: [Q] Factor
>  Date: 30 May 1995 01:28:03 GMT
>  
>  If I have a funktion like:
>  
>  f[x_] = (a + b x + c x^2) (2x - 2x^2)
>  
>  How can I factor it like that?
>  
>  f[x_] = a (2x-2x^2) + b x (2x-2x^2) + c x^2 (2x-2x^2) 
>            \_______/     \_________/     \___________/  
>                I              II              III
>  
>  And how can I get those factors?
>  
>  I    -> (2x-2x^2)
>  II   -> x (2x-2x^2)
>  III  -> x^2 (2x-2x^2)
>  
>  Answers are greatly appreciated.
>  
>  Bernd





  • Prev by Date: Evaluation of Special function: A bug?
  • Previous by thread: Re: [Q] Factor
  • Next by thread: Question about NDSolve and loops