Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.soft-sys.math.maple > #984

Re: factor

Show key headers only | View raw

On Thursday, October 16, 2014 4:36:46 PM UTC-4, acer wrote:
> On Thursday, October 16, 2014 11:34:48 AM UTC-4, William Unruh wrote:
> 
> > On 2014-10-16, maw...@gmail.com wrote:
> 
> > 
> 
> > > On Tuesday, October 14, 2014 9:15:35 PM UTC-4, maw...@gmail.com wrote:
> 
> > 
> 
> > >> Let p=3*x^3+25*x^2-10. How to compute a second-degree polynomial factor of this polynomial p in Maple?
> 
> > 
> 
> > >
> 
> > 
> 
> > > Still not worked out, the factor should be a polynomial of degree 2 with real coefficients. 
> 
> > 
> 
> > 
> 
> > 
> 
> > Since it must also have a first-degree polynomial factor , that would probably
> 
> > 
> 
> > be easier to find. which is x-.6105 approximately (done with maple but
> 
> > 
> 
> > probably not in a way you would want-- plotting it).Actually it has 3
> 
> > 
> 
> > real roots so there would be three second degree factors. 
> 
> > 
> 
> > Since this is a cubic you could also find the factors exactly (in terms
> 
> > 
> 
> > of cube roots).
> 
> 
> 
> He stated that he doesn't want I=sqrt(-1) to appear, and they may in the roots expressed in terms of radicals.
> 
> 
> 
> However, the roots for this example can be expressed in a mix or trig and arctrig without I appearing.
> 
> 
> 
> restart:                             
> 
> p := 3*x^3+25*x^2-10:                
> 
> R := map(simplify@evalc, [solve(p)]):
> 
> map(lprint, R):                      
> 
> 
> 
> -25/9+50/9*sin(1/3*arctan(9/2882*18021^(1/2))+1/6*Pi)
> 
> 
> 
> -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
> 
> (1/3*arctan(9/2882*3^(1/2)*6007^(1/2))+1/6*Pi)
> 
> 
> 
> -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
> 
> (1/3*arctan(9/2882*3^(1/2)*6007^(1/2))+1/6*Pi)
> 
> 
> 
> 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.

This is not what I wanted. Any resulting polynomial is still presented with transcendental numbers, not algebraic numbers. The original question is to ask if Maple is able to factor out a linear factor over algebraic number fields. Thanks!

Back to comp.soft-sys.math.maple | Previous | Next — Previous in thread | Next in thread | Find similar

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

csiph-web