Solving $x^5+x^4-12x^3-21x^2+x+5=0$

Question

$$x^5+x^4-12x^3-21x^2+x+5=0$$ I think it can be solved by trigonometric ways, but how?

Answer 1

Maple gives one of the roots as $$ 1/20\,\sqrt [5]{31}{4}^{4/5}\sqrt [5]{{\frac {125\,\sqrt {5}\sqrt {-65 -22\,\sqrt {5}}-409\,\sqrt {-65-22\,\sqrt {5}}-1575\,\sqrt {5}-3725}{ \sqrt {-65-22\,\sqrt {5}}}}}+{\frac { \left( 1577\,{\frac {\sqrt {5}}{ \sqrt {-65-22\,\sqrt {5}}}}-65\,\sqrt {5}+181\,\sqrt {-65-22\,\sqrt {5 }}-13 \right) {31}^{2/5}{4}^{3/5}}{260\, \left( {\frac {125\,\sqrt {5} \sqrt {-65-22\,\sqrt {5}}-409\,\sqrt {-65-22\,\sqrt {5}}-1575\,\sqrt { 5}-3725}{\sqrt {-65-22\,\sqrt {5}}}} \right) ^{3/5}}}+{\frac { \left( 437\,{\frac {\sqrt {5}}{\sqrt {-65-22\,\sqrt {5}}}}+65\,\sqrt {5}+11\, \sqrt {-65-22\,\sqrt {5}}+143 \right) {31}^{3/5}{4}^{2/5}}{260\, \left( {\frac {125\,\sqrt {5}\sqrt {-65-22\,\sqrt {5}}-409\,\sqrt {- 65-22\,\sqrt {5}}-1575\,\sqrt {5}-3725}{\sqrt {-65-22\,\sqrt {5}}}} \right) ^{2/5}}}+1/5\,{\frac {{31}^{4/5}\sqrt [5]{4}}{\sqrt [5]{{ \frac {125\,\sqrt {5}\sqrt {-65-22\,\sqrt {5}}-409\,\sqrt {-65-22\, \sqrt {5}}-1575\,\sqrt {5}-3725}{\sqrt {-65-22\,\sqrt {5}}}}}}}-1/5 $$ (the others are similar expressions)

Answer 2

Inspecting the discriminant and the small primes where the polynomial splits suggests that its splitting field is the class field of modulus $31$ corresponding to the subgroup $\{\pm 1, \pm 5, \pm 6\}$ of $(\Bbb Z/31\Bbb Z)/\{\pm 1\}$.

Which means that the splitting field is $\Bbb Q\left(2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)\right)$.

Then I computed the minimal polynomial of this quantity (pure curiosity), and ended up on your polynomial exactly.

So the roots are $2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)$, and its conjugates $$2\left(\cos\frac{4\pi}{31}+\cos\frac{20\pi}{31}+\cos\frac{24\pi}{31}\right) , \\ 2\left(\cos\frac{8\pi}{31}+\cos\frac{22\pi}{31}+\cos\frac{14\pi}{31}\right), \\ 2\left(\cos\frac{16\pi}{31}+\cos\frac{18\pi}{31}+\cos\frac{28\pi}{31}\right),\\ 2\left(\cos\frac{30\pi}{31}+\cos\frac{26\pi}{31}+\cos\frac{6\pi}{31}\right).$$

I have no idea on how to guess this without the heavy machinery.

Answer 3

For primes $p \equiv 1 \pmod {10},$ the method of Gauss gives the roots of a certain quintic as sums of $\zeta^n,$ where $\zeta = e^{2 \pi i/p}.$ In turn, the collection of exponents $n$ is of the form $w^k.$ Example, for $p = 31$ the exponents are $6^k$ for $0 \leq k \leq 5.$ Here are the first 35 such polynomials and the exponents for one of the roots. Compare with the list for $p = 11, 31, 41, 61, 71$ in https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics

  x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1   prime  11 zeta  exponents 10^k 
  x^5 + x^4 - 12 x^3 - 21 x^2 + 1 x + 5   prime  31 zeta  exponents 6^k 
  x^5 + x^4 - 16 x^3 + 5 x^2 + 21 x - 9   prime  41 zeta  exponents 3^k 
  x^5 + x^4 - 24 x^3 - 17 x^2 + 41 x - 13   prime  61 zeta  exponents 21^k 
  x^5 + x^4 - 28 x^3 + 37 x^2 + 25 x + 1   prime  71 zeta  exponents 23^k 
  x^5 + x^4 - 40 x^3 + 93 x^2 - 21 x - 17   prime  101 zeta  exponents 32^k 
  x^5 + x^4 - 52 x^3 - 89 x^2 + 109 x + 193   prime  131 zeta  exponents 18^k 
  x^5 + x^4 - 60 x^3 - 12 x^2 + 784 x + 128   prime  151 zeta  exponents 23^k 
  x^5 + x^4 - 72 x^3 - 123 x^2 + 223 x - 49   prime  181 zeta  exponents 17^k 
  x^5 + x^4 - 76 x^3 - 359 x^2 - 437 x - 155   prime  191 zeta  exponents 11^k 
  x^5 + x^4 - 84 x^3 - 59 x^2 + 1661 x + 269   prime  211 zeta  exponents 26^k 
  x^5 + x^4 - 96 x^3 - 212 x^2 + 1232 x + 512   prime  241 zeta  exponents 11^k 
  x^5 + x^4 - 100 x^3 - 20 x^2 + 1504 x + 1024   prime  251 zeta  exponents 2^k 
  x^5 + x^4 - 108 x^3 - 401 x^2 - 13 x + 845   prime  271 zeta  exponents 12^k 
  x^5 + x^4 - 112 x^3 - 191 x^2 + 2257 x + 967   prime  281 zeta  exponents 6^k 
  x^5 + x^4 - 124 x^3 + 535 x^2 - 413 x - 539   prime  311 zeta  exponents 11^k 
  x^5 + x^4 - 132 x^3 - 887 x^2 - 1843 x - 1027   prime  331 zeta  exponents 13^k 
  x^5 + x^4 - 160 x^3 + 369 x^2 + 879 x - 29   prime  401 zeta  exponents 26^k 
  x^5 + x^4 - 168 x^3 + 219 x^2 + 3853 x - 3517   prime  421 zeta  exponents 32^k 
  x^5 + x^4 - 184 x^3 - 129 x^2 + 4551 x + 5419   prime  461 zeta  exponents 13^k 
  x^5 + x^4 - 196 x^3 + 59 x^2 + 2019 x + 1377   prime  491 zeta  exponents 32^k 
  x^5 + x^4 - 208 x^3 - 771 x^2 + 4143 x + 2083   prime  521 zeta  exponents 24^k 
  x^5 + x^4 - 216 x^3 + 1147 x^2 - 805 x - 2629   prime  541 zeta  exponents 11^k 
  x^5 + x^4 - 228 x^3 + 868 x^2 + 3056 x - 7552   prime  571 zeta  exponents 2^k 
  x^5 + x^4 - 240 x^3 + 1755 x^2 - 3731 x + 2399   prime  601 zeta  exponents 17^k 
  x^5 + x^4 - 252 x^3 + 2095 x^2 - 5785 x + 5069   prime  631 zeta  exponents 24^k 
  x^5 + x^4 - 256 x^3 - 564 x^2 + 5328 x - 5120   prime  641 zeta  exponents 21^k 
  x^5 + x^4 - 264 x^3 - 185 x^2 + 16837 x + 4851   prime  661 zeta  exponents 32^k 
  x^5 + x^4 - 276 x^3 - 1299 x^2 + 5329 x + 15581   prime  691 zeta  exponents 11^k 
  x^5 + x^4 - 280 x^3 + 2047 x^2 - 3791 x + 1699   prime  701 zeta  exponents 23^k 
  x^5 + x^4 - 300 x^3 - 2313 x^2 - 3761 x - 571   prime  751 zeta  exponents 11^k 
  x^5 + x^4 - 304 x^3 + 2831 x^2 - 8925 x + 8775   prime  761 zeta  exponents 3^k 
  x^5 + x^4 - 324 x^3 - 3471 x^2 - 12431 x - 13603   prime  811 zeta  exponents 12^k 
  x^5 + x^4 - 328 x^3 - 1215 x^2 + 3573 x + 2179   prime  821 zeta  exponents 32^k 
  x^5 + x^4 - 352 x^3 - 2361 x^2 + 4257 x + 9967   prime  881 zeta  exponents 29^k 

Answer 4

This is an old question, but should be interesting to answer. If you want an aesthetic version, then the solution to,

$$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$

is given by,

$$x = \frac{1}{5}\left(-1+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$

and the $z_i$ are the roots of the quartic,

$$z^4 + 12679 z^3 + 78678031 z^2 + 362989005529 z + 31^{10} = 0$$

Notice how the quartic has a constant term that is a fifth power. This factors over the extension $\sqrt{5}$. So alternatively,

$$x = \frac{1}{5}\left(\beta_0-1+\frac{31}{\beta_0}+\frac{31}{\alpha}+\alpha\right)$$

where,

$$\alpha = \left(\tfrac{31}{4}\right)^{1/5}\left(-409-125\sqrt{5}+5\sqrt{-10(925-409\sqrt{5})}\right)^{1/5}\tag1$$

$$\beta_0 = \left(\tfrac{31}{4}\right)^{1/5}\left(-409+125\sqrt{5}+5\sqrt{-10(925+409\sqrt{5})}\right)^{1/5}\tag2$$

Added:

Israel's Maple answer can be simplified. The five roots $x_k$ for $k=0,1,2,3,4$ are,

$$x_k = \frac{1}{5}\left(\frac{1}{\beta_k^{-1}}-\frac{1}{\beta_k^0}+\frac{31}{\beta_k^1}+\frac{31a}{\beta_k^2}+\frac{31b}{\beta_k^3} \right)$$

where,

$$a=\frac{\beta_0^2}{\alpha}=\tfrac{1}{4}\left(11+5\sqrt{5}+\sqrt{-10(25-11\sqrt{5})}\right)$$

$$b=\frac{\alpha\,\beta_0^3}{31}=\tfrac{1}{4}\left(-1-5\sqrt{5}+\sqrt{-10(1525-\sqrt{5})}\right)$$

$$\beta_k = e^{2\pi\,i\,k/5}\,\left(31ab\right)^{1/5}\tag3$$

So it turns out that the root $(2)$ can be factored as $(3)$. Maple's answer also has the advantage that only one fifth root extraction of a complex number is needed.