Re: Re: Re: Poles and Complex Factoring

*To*: mathgroup at smc.vnet.net*Subject*: [mg52221] Re: [mg52216] Re: [mg52176] Re: [mg6011] Poles and Complex Factoring*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 15 Nov 2004 03:17:28 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Another way: In[1]:= SolveAlways[x^2 + 2*x + 10 == (x + c)*(x + d), x] Out[1]= {{c -> 1 - 3*I, d -> 1 + 3*I}, {c -> 1 + 3*I, d -> 1 - 3*I}} In[2]:= Union[(x + c)*(x + d) /. %] Out[2]= {(x + (1 - 3*I))*(x + (1 + 3*I))} Andrzej Kozlowski On 15 Nov 2004, at 10:15, Bob Hanlon wrote: > 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: [mg52221] [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. >> >> >