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MathGroup Archive 2004

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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.
>>
>>
>


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