Re: Factoring

*To*: mathgroup at smc.vnet.net*Subject*: [mg65128] Re: [mg65069] Factoring*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Wed, 15 Mar 2006 06:29:03 -0500 (EST)*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst*References*: <200603141059.FAA24051@smc.vnet.net>*Reply-to*: murray at math.umass.edu*Sender*: owner-wri-mathgroup at wolfram.com

In your first expression, I cannot see what the symbols are after the "2*" inside the Sin function. To my eye it looks like it might be a "hatted" I followed by a period, but that makes no sense. If that is supposed to be an I also, then expression you gave is real-valued (assuming that r is real): 4 I r^2 Sin[2 I] + I r^5 Sin[5 I -4*r^2*Sinh[2] - r^5*Sinh[5] If you prefer: TrigToExp[4 I r^2 Sin[2 I] + I r^5 Sin[5 I]] (2*r^2)/E^2 - 2*E^2*r^2 + r^5/(2*E^5) - (E^5*r^5)/2 In any event, for the last expression you gave, you could use: 4x I Sin[t] + 28x^3 I Cos[t] I*(28*x^3*Cos[t] + 4*x*Sin[t]) (I've converted the output cells to InputForm.) I realize that the above does not respond to your more general question. Matt wrote: > Hello Mathgroup, > I'm sure that I've overlooked something obvious, but for the past two > and a half hours, I've been trying to figure out how to use built-in > Mathematica functions to just factor the imaginary number 'I' out of > this: > > 4*I*r^2*Sin[2*Î¸] + I*r^5*Sin[5*Î¸] > > using Factor[] gives me I*r^2*(4*Sin[2*Î¸] + r^3*Sin[5*Î¸]) > > using FactorTerms[] gives me I*(4*r^2*Sin[2*Î¸] + r^5*Sin[5*Î¸]) which > is what I want, but as soon as I add in a common numerical factor, it > also factors that out as well, > > e.g. 4*I*r^2*Sin[2*Î¸] + 8*I*r^5*Sin[5*Î¸] > > I finally just decided to divide the whole thing by 'I', and do some > reconstruction of my expression, but I have the distinct feeling I've > totally missed something very simple. As an additional example, what > if I just wanted to extract 2*x*I out of the following: > > 4*x*I*Sin[t] + 28*x^3*I*Cos[t] > > how would I do that? In general, I'm looking for a function that says > "Given an expression 'expr' and another expression 'sub' that is common > to all additive terms of 'expr', give a result that is the product of > 'sub' and the result of factoring 'sub' out of 'expr'". > > Thanks, > > Matt > > -- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305

**References**:**Factoring***From:*"Matt" <anonmous69@netscape.net>