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I want to find the analytical roots of this polynomial

- x - a^3*x*(a^2*x*(a*x*(x - 1) + 1)*(x - 1) + 1)*(a*x*(x - 1) + 1)*(x - 1)

Solve gives the first 2 roots as x=0 and x=1-1/a while the other 6 roots - as symbolic expressions using Root object. Is there a way to find analytical solutions for other 6 roots expressed without Root object?

Pavlo Fesenko
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Yogesh Yadav
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    The other six roots apparently do not exist in closed form. – bbgodfrey Jan 30 '16 at 14:33
  • Solve[-x - a^3*x*(a^2*x*(a*x*(x - 1) + 1)*(x - 1) + 1)*(a*x*(x - 1) + 1)*(x - 1) == 0, x] works for me: http://i.stack.imgur.com/SULgx.png – Michael E2 Jan 30 '16 at 14:33
  • I am very curious how did you find the numerical roots? Your a is not defined. – Pavlo Fesenko Jan 30 '16 at 20:41
  • Could use Roots[-x - a^3*x*(a^2*x*(a*x*(x - 1) + 1)*(x - 1) + 1)*(a*x*(x - 1) + 1)*(x - 1) == 0, x] which of course is a documented function. – Daniel Lichtblau Jan 30 '16 at 22:18
  • Are you interested only in real solutions or also in imaginary ones? I think your problem will be much simpler if you are only looking for real solutions. – Pavlo Fesenko Feb 03 '16 at 11:17
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    @PavloFesenko Note that it is an error to say Solve returns numerical solutions. It returns symbolic ones for the other 6 roots in terms of Root (see for instance http://mathematica.stackexchange.com/a/13768). If it is returning numerical solutions for you, it is a bug. – Michael E2 Feb 03 '16 at 13:29
  • @MichaelE2 Thanks for pointing it out. I have corrected my edit. – Pavlo Fesenko Feb 03 '16 at 14:17

1 Answers1

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You have the irreducible polynomial of the 7th degree:

-x + a^3 x - a^3 x^2 - a^4 x^2 - a^5 x^2 + 2 a^4 x^3 + 2 a^5 x^3 + 
 2 a^6 x^3 - a^4 x^4 - a^5 x^4 - 6 a^6 x^4 - a^7 x^4 + 6 a^6 x^5 + 
 4 a^7 x^5 - 2 a^6 x^6 - 6 a^7 x^6 + 4 a^7 x^7 - a^7 x^8

The general algebraic solution doesn't exist for polynomials of the 5th degree and higher with arbitrary coefficients. This is called Abel-Ruffini theorem.

mmal
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Pavlo Fesenko
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    Some higher order polynomials can be solved analytically eg x^8-1==0. Abel-Ruffini says there's no general algorithm in terms of radicals. – alancalvitti Jan 30 '16 at 14:49