Re: simplify polynomial
- To: mathgroup at smc.vnet.net
- Subject: [mg89196] Re: simplify polynomial
- From: dh <dh at metrohm.ch>
- Date: Thu, 29 May 2008 07:02:46 -0400 (EDT)
- References: <g1j6im$qop$1@smc.vnet.net>
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. Oberdorfstr. 68 CH-9100 Herisau Tel. +41 71 353 8585, Fax +41 71 353 8907 E-Mail:<mailto:dh at metrohm.com> Internet:<http://www.metrohm.com>