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Groups > comp.soft-sys.math.maple > #984
| Newsgroups | comp.soft-sys.math.maple |
|---|---|
| Date | 2014-10-16 19:45 -0700 |
| References | <fc590dca-6f8f-435e-8f9e-b47741417775@googlegroups.com> <ab7ef8a3-5143-4369-a0f1-8e2435b2e5c8@googlegroups.com> <m1ooin$3jd$1@dont-email.me> <254ca473-4f4f-4814-a2a9-4ef0f62efac9@googlegroups.com> |
| Message-ID | <e8d8a56e-441a-4dde-9439-9bcf4067f46b@googlegroups.com> (permalink) |
| Subject | Re: factor |
| From | mawxfl@gmail.com |
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!
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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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