Calculator that can factor and find exact roots of polynomials - Page 4

alpertron · 2020-11-30, 17:15

I should add TEX output to my calculators when requested by user. That would be an interesting addition to the programs.

alpertron · 2020-12-13, 21:06

I've just added TeX output to my polynomial factorization calculator located at https://www.alpertron.com.ar/POLFACT.HTM

For example, the roots of x¹⁷ + 1 are:

[math]\begin{array}{l}
\bullet\,\,x_{1} = -1\\
\bullet\,\,x_{2} = \cos{\frac{\pi }{17}} + i \sin{\frac{\pi }{17}} = \frac{1}{16}\left(1-\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{3} = \cos{ \frac{3\pi }{17}} + i \sin{\frac{3\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{4} = \cos{ \frac{5\pi }{17}} + i \sin{\frac{5\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}-2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{5} = \cos{ \frac{7\pi }{17}} + i \sin{\frac{7\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}-\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{6} = \cos{ \frac{9\pi }{17}} + i \sin{\frac{9\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}-2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{7} = \cos{ \frac{11\pi }{17}} + i \sin{\frac{11\pi }{17}} = -\frac{1}{16}\left(-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{8} = \cos{ \frac{13\pi }{17}} + i \sin{\frac{13\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}-\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{9} = \cos{ \frac{15\pi }{17}} + i \sin{\frac{15\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{10} = \cos{ \frac{19\pi }{17}} + i \sin{\frac{19\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{11} = \cos{ \frac{21\pi }{17}} + i \sin{\frac{21\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}-\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{12} = \cos{ \frac{23\pi }{17}} + i \sin{\frac{23\pi }{17}} = -\frac{1}{16}\left(-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{13} = \cos{ \frac{25\pi }{17}} + i \sin{\frac{25\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}-2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\
\bullet\,\,x_{14} = \cos{ \frac{27\pi }{17}} + i \sin{\frac{27\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}-\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{15} = \cos{ \frac{29\pi }{17}} + i \sin{\frac{29\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}-2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{16} = \cos{ \frac{31\pi }{17}} + i \sin{\frac{31\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\
\bullet\,\,x_{17} = \cos{ \frac{33\pi }{17}} + i \sin{\frac{33\pi }{17}} = \frac{1}{16}\left(1-\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\
\end{array}[/math]

I had to change a lot of code to do this, so it is possible that there are some errors. So I will appreciate if you post here the error(s) you can find.

alpertron · 2020-12-25, 18:21

I've just added FFT for modular polynomial multiplications when the modulus is small. This enables faster factoring when trying to factor integer polynomials, especially when the number of modular factors is small.

For example, the time required for indicating that x⁹⁰⁰ + 2 is irreducible changed from 12 seconds to 4.7 seconds.

At this moment the bottleneck is the division of polynomials. It appears that I have to implement a divide-and-conquer approach.

R. Gerbicz · 2020-12-25, 18:54

Quote:

Originally Posted by alpertron

For example, the time required for indicating that x⁹⁰⁰ + 2 is irreducible changed from 12 seconds to 4.7 seconds.

That is way slower than my brain, it is irreducible with Eisenstein's criterion using p=2, ref:
https://en.wikipedia.org/wiki/Eisenstein%27s_criterion

alpertron · 2020-12-25, 19:05

You are right. But the Eisenstein criterion cannot be used for all polynomials.

There is still more room for optimization.

alpertron · 2021-03-02, 01:53

Quote:

Originally Posted by R. Gerbicz

Try x^720-1. On an older Chrome it is just crashing, on Firefox after some computation it has given: "index out of bounds", the 2nd run is still computing and it is in the: "Computing LLL in matrix of 43 × 43."

There were several buffer overflows that I've just fixed during this week. Now it works.

alpertron · 2022-06-02, 12:21

I've just uploaded to the Web server a new version of the polynomial root calculator. Now it can compute the exact roots of quintic equations when the Galois group is 5 (all five roots are real). In this case trigonometric functions are required if we do not want to use complex numbers.

Example:

$x^{5} + x^{4} - 4x^{3} - 3x^{2} + 3x + 1 = 0$

$\bullet\,\,x_{1} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right) + \cos{\left(\frac{1}{5}\left(4\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$
$\bullet\,\,x_{2} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(2\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)$
$\bullet\,\,x_{3} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(4\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\left(6\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$
$\bullet\,\,x_{4} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(6\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\left(2\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$
$\bullet\,\,x_{5} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(8\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\left(8\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$

You can select Pari-GP output and check using that tool that the formulas generated by the solver are correct.

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2020-11-30, 17:15	#34
alpertron Aug 2002 Buenos Aires, Argentina 2²×3×11² Posts	I should add TEX output to my calculators when requested by user. That would be an interesting addition to the programs.

2020-12-13, 21:06	#35
alpertron Aug 2002 Buenos Aires, Argentina 2²·3·11² Posts	I've just added TeX output to my polynomial factorization calculator located at https://www.alpertron.com.ar/POLFACT.HTM For example, the roots of x¹⁷ + 1 are: [math]\begin{array}{l} \bullet\,\,x_{1} = -1\\ \bullet\,\,x_{2} = \cos{\frac{\pi }{17}} + i \sin{\frac{\pi }{17}} = \frac{1}{16}\left(1-\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{3} = \cos{ \frac{3\pi }{17}} + i \sin{\frac{3\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{4} = \cos{ \frac{5\pi }{17}} + i \sin{\frac{5\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}-2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{5} = \cos{ \frac{7\pi }{17}} + i \sin{\frac{7\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}-\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{6} = \cos{ \frac{9\pi }{17}} + i \sin{\frac{9\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}-2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{7} = \cos{ \frac{11\pi }{17}} + i \sin{\frac{11\pi }{17}} = -\frac{1}{16}\left(-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{8} = \cos{ \frac{13\pi }{17}} + i \sin{\frac{13\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}-\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{9} = \cos{ \frac{15\pi }{17}} + i \sin{\frac{15\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right) + \frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{10} = \cos{ \frac{19\pi }{17}} + i \sin{\frac{19\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{11} = \cos{ \frac{21\pi }{17}} + i \sin{\frac{21\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}-\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{12} = \cos{ \frac{23\pi }{17}} + i \sin{\frac{23\pi }{17}} = -\frac{1}{16}\left(-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{13} = \cos{ \frac{25\pi }{17}} + i \sin{\frac{25\pi }{17}} = -\frac{1}{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}-2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}+2\sqrt{34-2\sqrt{17}}+4\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}}\\ \bullet\,\,x_{14} = \cos{ \frac{27\pi }{17}} + i \sin{\frac{27\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}-\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}+2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{15} = \cos{ \frac{29\pi }{17}} + i \sin{\frac{29\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}-2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}+4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{16} = \cos{ \frac{31\pi }{17}} + i \sin{\frac{31\pi }{17}} = \frac{1}{16}\left(1+\sqrt{17}+\sqrt{34+2\sqrt{17}}+2\sqrt{17-3\sqrt{17}-\sqrt{170-38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34+2\sqrt{17}-2\sqrt{34+2\sqrt{17}}-4\sqrt{17-3\sqrt{17}+\sqrt{170-38\sqrt{17}}}}\\ \bullet\,\,x_{17} = \cos{ \frac{33\pi }{17}} + i \sin{\frac{33\pi }{17}} = \frac{1}{16}\left(1-\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}+\sqrt{170+38\sqrt{17}}}\right)-\frac{i}{8}\sqrt{34-2\sqrt{17}-2\sqrt{34-2\sqrt{17}}-4\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\\ \end{array}[/math] I had to change a lot of code to do this, so it is possible that there are some errors. So I will appreciate if you post here the error(s) you can find.

2020-12-25, 18:21	#36
alpertron Aug 2002 Buenos Aires, Argentina 2²·3·11² Posts	I've just added FFT for modular polynomial multiplications when the modulus is small. This enables faster factoring when trying to factor integer polynomials, especially when the number of modular factors is small. For example, the time required for indicating that x⁹⁰⁰ + 2 is irreducible changed from 12 seconds to 4.7 seconds. At this moment the bottleneck is the division of polynomials. It appears that I have to implement a divide-and-conquer approach.

2020-12-25, 19:05	#38
alpertron Aug 2002 Buenos Aires, Argentina 2²×3×11² Posts	You are right. But the Eisenstein criterion cannot be used for all polynomials. There is still more room for optimization.

2022-06-02, 12:21	#40
alpertron Aug 2002 Buenos Aires, Argentina 2²×3×11² Posts	I've just uploaded to the Web server a new version of the polynomial root calculator. Now it can compute the exact roots of quintic equations when the Galois group is 5 (all five roots are real). In this case trigonometric functions are required if we do not want to use complex numbers. Example: $x^{5} + x^{4} - 4x^{3} - 3x^{2} + 3x + 1 = 0$ $\bullet\,\,x_{1} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right) + \cos{\left(\frac{1}{5}\left(4\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$ $\bullet\,\,x_{2} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(2\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)$ $\bullet\,\,x_{3} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(4\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\left(6\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$ $\bullet\,\,x_{4} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(6\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\left(2\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$ $\bullet\,\,x_{5} = \frac{1}{-5} + \frac{2}{5}\sqrt{11}\left(\cos{\left(\frac{1}{5}\left(8\pi + \arccos\left(\frac{-89}{484}\sqrt{11} + \frac{25}{484}\sqrt{55}\right)\right)\right) + \cos{\left(\frac{1}{5}\left(8\pi + \arccos\left(\frac{-89}{484}\sqrt{11} - \frac{25}{484}\sqrt{55}\right)\right)\right)\right)$ You can select Pari-GP output and check using that tool that the formulas generated by the solver are correct.