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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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