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Re: factor

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Date	Fri, 17 Oct 2014 10:39:23 -0700 (PDT)
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Subject	Re: factor
From	acer <maple@rogers.com>
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Xref	csiph.com comp.soft-sys.math.maple:987

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On Friday, October 17, 2014 1:33:39 AM UTC-4, William Unruh wrote:
> Please stop using googlegroups to post. It inserts a blank line between
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> every line of quoted text, which causes and exponential runaway.
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> 
> 
> On 2014-10-17, acer <m...@rogers.com> wrote:
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> > Le jeudi 16 octobre 2014 16:36:46 UTC-4, acer a ?crit?:
> 
> >> On Thursday, October 16, 2014 11:34:48 AM UTC-4, William Unruh wrote:
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> >> 
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> >> > On 2014-10-16, maw...@gmail.com wrote:
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> >> 
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> >> > 
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> >> > > On Tuesday, October 14, 2014 9:15:35 PM UTC-4, maw...@gmail.com wrote:
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> >> 
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> >> > 
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> >> 
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> >> > >> Let p=3*x^3+25*x^2-10. How to compute a second-degree polynomial factor of this polynomial p in Maple?
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> >> > > Still not worked out, the factor should be a polynomial of degree 2 with real coefficients. 
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> >> > Since it must also have a first-degree polynomial factor , that would probably
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> >> > be easier to find. which is x-.6105 approximately (done with maple but
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> >> > probably not in a way you would want-- plotting it).Actually it has 3
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> >> > real roots so there would be three second degree factors. 
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> >> > Since this is a cubic you could also find the factors exactly (in terms
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> >> > of cube roots).
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> >> He stated that he doesn't want I=sqrt(-1) to appear, and they may in the roots expressed in terms of radicals.
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> In this case all roots are real. Plot it and see.


We were all aware of this already, I'm sure. I certainly was.

 
> >> However, the roots for this example can be expressed in a mix or trig and arctrig without I appearing.
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> >> restart:                             
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> >> 
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> >> p := 3*x^3+25*x^2-10:                
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> >> 
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> >> R := map(simplify@evalc, [solve(p)]):
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> >> 
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> >> map(lprint, R):                      
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> >> 
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> >> 
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> >> 
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> >> -25/9+50/9*sin(1/3*arctan(9/2882*18021^(1/2))+1/6*Pi)
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> >> 
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> >> 
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> >> 
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> >> -25/9-25/9*3^(1/2)*sin(-1/3*arctan(9/2882*3^(1/2)*6007^(1/2))+1/3*Pi)-25/9*sin
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> >> 
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> >> (1/3*arctan(9/2882*3^(1/2)*6007^(1/2))+1/6*Pi)
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> >> 
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> >> 
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> >> 
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> >> -25/9+25/9*3^(1/2)*sin(-1/3*arctan(9/2882*3^(1/2)*6007^(1/2))+1/3*Pi)-25/9*sin
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> >> 
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> >> (1/3*arctan(9/2882*3^(1/2)*6007^(1/2))+1/6*Pi)
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> >> 
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> >> 
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> >> 
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> >> So, as you suggest, any of the 6 possible degree 2 factors can be had just by using any pair of the three roots. In my earlier answer I did just that, and chose the 2nd and 3rd roots returned by the `solve` command.
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> >
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> So did I in my numerical solution to the problem.
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> >
> 
> > I meant any of the 3 possible degree 2 factors, not 6, sorry. Long day.

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Thread

factor mawxfl@gmail.com - 2014-10-14 18:15 -0700
  Re: factor "Nasser M. Abbasi" <nma@12000.org> - 2014-10-14 21:42 -0500
  Re: factor acer <maple@rogers.com> - 2014-10-14 21:13 -0700
  Re: factor mawxfl@gmail.com - 2014-10-15 21:09 -0700
    Re: factor acer <maple@rogers.com> - 2014-10-15 22:17 -0700
    Re: factor William Unruh <unruh@invalid.ca> - 2014-10-16 15:34 +0000
      Re: factor acer <maple@rogers.com> - 2014-10-16 13:36 -0700
        Re: factor acer <maple@rogers.com> - 2014-10-16 17:55 -0700
          Re: factor William Unruh <unruh@invalid.ca> - 2014-10-17 05:33 +0000
            Re: factor acer <maple@rogers.com> - 2014-10-17 10:39 -0700
        Re: factor mawxfl@gmail.com - 2014-10-16 19:45 -0700
          Re: factor William Unruh <unruh@invalid.ca> - 2014-10-17 05:35 +0000
            Re: factor mawxfl@gmail.com - 2014-10-17 14:33 -0700
              Re: factor "G. A. Edgar" <edgar@math.ohio-state.edu.invalid> - 2014-10-19 06:42 -0600
  Re: factor Mario Lemelin <mario.lemelin@cgocable.ca> - 2014-10-20 09:08 -0700

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