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MathGroup Archive 2004

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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.
> 
> 


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