[mg89196] Re: simplify polynomial

[dh <dh at metrohm.ch>]

Hi,

do not use funny characters like: λ, I replace it b:x.

It is not very clear to me what you mean by incomplete factorization. 

Well, a polynomial (with variable x) can always be factorized, using its 

zeros.In your case of a third degree poly: c0 (x-c1)(x-c2)(x-c3). If you 

want to split an additional summand, you could eliminate e.g. x1 and add 

the summand: c0 x(x-c2)(x-c3)+c4. The ci you can get using SolveAlways:

SolveAlways[yourPoly==c0 x(x-c2)(x-c3)+c4,x]

will give you the ci.

hope this helps, Daniel



vmarian wrote:

> I want to simplify a polynomial by factorizing the most of the  terms as possible(that means a product of terms plus a small other 

> number of terms). I didn't find how to do such incomplete factorization. Is there a command to do it. Factor works only for complete factorizations. 

> 

> \!\(Simplify[2 - λ - λ\^2 - 3\ hmb\ \((\(-2\) + λ + 

>       λ\^2)\) + hmb\^3\ \((5 - 12\ λ + 4\ λ\^2)\) + hmb\^2\ \((6 - 3\ λ - 

>       8\ λ\^2 + 2\ λ\^3)\)]\)

> 

> Thank you.

> 





-- 



Daniel Huber

Metrohm Ltd.

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