Re: Re: Poles and Complex Factoring

*To*: mathgroup at smc.vnet.net*Subject*: [mg52216] Re: [mg52176] Re: [mg6011] Poles and Complex Factoring*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sun, 14 Nov 2004 20:15:18 -0500 (EST)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

Your solution does not check out (x-2-Sqrt[10])(x-2+Sqrt[10])//ExpandAll -6 - 4*x + x^2 (-b + Sqrt[b^2-4*a*c])/(2a) /. {a->1,b->2,c->10} -1 + 3*I Using Factor as previously posted by several posters: Factor[x^2+2x+10,GaussianIntegers->True] ((1 - 3*I) + x)*((1 + 3*I) + x) Or equating terms: Union[(x-a)(x-b) /. Solve[Thread[CoefficientList[#, x]&/@ (x^2 + 2x + 10 == (x-a)(x-b))], {a,b}]] {((1 - 3*I) + x)*((1 + 3*I) + x)} Bob Hanlon > > From: godfreyau2003 at hotmail.com (Godfrey) To: mathgroup at smc.vnet.net > Date: 2004/11/14 Sun AM 04:30:12 EST > To: mathgroup at smc.vnet.net > Subject: [mg52216] [mg52176] Re: [mg6011] Poles and Complex Factoring > > On 11 Feb 1997 01:29:40 -0500, peter wrote: > >Dear All, > > > >I know how to calculate the residue of a fuction using Mathematica, but how > can I > >use Mathematica to calculate the order of a complex pole? > > > >It would also be nice for Mathematica to tell me if a particular singularity > is an > >essential singularity, removable singularity or a pole...but this is > not > >necessary; just icing on the cake. > > > >Also, is there a way to factor polynomials with imaginary roots? > >Something like: > > > > Factor[ x^2 + 2x + 10 ] = (x - 1 + 4.5 I)(x - 1 - 4.5 I) > > > >Much thanks in advance! > > > >Peter > > > >-- > >Birthdays are good for you: A federal funded project has recently > determined > >that people with the most number of birthdays will live the > longest..... > >-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+-=><=-+-+- =><=-+-+-=><=-+-+-=><=-+-+-=><=- > > I BOYCOTT ANY COMPANY THAT USES MASS ADVERTISING ON THE INTERNET > > Let x^2 + 2x + 10 = (x-a)(x-b) > Obviously, a and b are the roots of the equation x^2 + 2x + 10 = 0 > ------ (*) > By quadratic forumula, we can find that the roots of (*)=2+sqrt(10) or > 2-sqrt(10) > Therefore, > x^2 + 2x + 10 = {x-[2+sqrt(10)]}{x-[2-sqrt(10)]} > ie. x^2 + 2x + 10 = [x-2-sqrt(10)][x-2+sqrt(10)] > > Factorization of other polynomials with imaginary roots can be done in > the same way. > >