Solving for the roots of a solvable quintic

In the Wikipedia page about quintics, there was a list of quintics that could be solved with trigonometric roots.

For example:x5+x4−4×3−3×2+3x+1x^5+x^4-4x^3-3x^2+3x+1\tag1 has roots of the form 2cos2kπ112\cos \frac {2k\pi}{11}
x^5+x^4-16x^3+5x^2+21x-9=0\tag2x^5+x^4-16x^3+5x^2+21x-9=0\tag2 has roots of the form \sum_{k=0}^{7}e^{\frac {2\pi i 3^k}{41}}\sum_{k=0}^{7}e^{\frac {2\pi i 3^k}{41}}

Question: Is there a formula or underlying structure to find the roots of the quintics?

Wikipedia says that the roots are the sums of the first nn-th roots of unity with n=10k+1n=10k+1, so I’m guessing we first have to find nn, or do we first have to find kk?

Any help is appreciated.

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en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem
– Jack D’Aurizio
Oct 21 at 1:43

1

Those polynomials are very special cases associated with Galois extensions of \mathbb{Q}\mathbb{Q} with degree 55. In general, the roots of a quintic equation cannot be found in terms of elementary functions.
– Jack D’Aurizio
Oct 21 at 1:45

But, for instance, any equation of the form \left(\frac{x+a}{x+b}\right)^6 = 1\left(\frac{x+a}{x+b}\right)^6 = 1, that is associated with the roots of a quintic, can be easily solved.
– Jack D’Aurizio
Oct 21 at 1:47

To recognize a solvable quintic from a non-solvable one from its coefficients is a quite involved problem in Galois (and symmetric functions) theory.
– Jack D’Aurizio
Oct 21 at 1:49

@JackD’Aurizio But is there a way to solve the quintic I specified in my question? I am interested in how they managed to produce such “complex” results.
– Frank
Oct 21 at 1:50

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ADDED: I remembered where I had seen this before, with lower degree; let \alpha \neq 1\alpha \neq 1 be a seventh root of unity, \alpha^7 = 1, \alpha^7 = 1, and take
\gamma = \alpha + \alpha^6, \gamma = \alpha + \alpha^6,
\gamma^2 = \alpha^2 + 2 + \alpha^5, \gamma^2 = \alpha^2 + 2 + \alpha^5,
\gamma^3 = \alpha^3 + 3 \alpha + 3 \alpha^6 + \alpha^4. \gamma^3 = \alpha^3 + 3 \alpha + 3 \alpha^6 + \alpha^4.
Therefore
\gamma^3 + \gamma^2 – 2 \gamma – 1 = \alpha^3 + \alpha^2 + \alpha + 1 + \alpha^6 + \alpha^5 + \alpha^4 = 0 \gamma^3 + \gamma^2 – 2 \gamma – 1 = \alpha^3 + \alpha^2 + \alpha + 1 + \alpha^6 + \alpha^5 + \alpha^4 = 0
This is a cubic with three irrational roots, and is https://en.wikipedia.org/wiki/Casus_irreducibilis

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I simply checked the one that seems easiest. Cute. Let \omega \neq 1\omega \neq 1 be an 11th root of unity, \omega^{11} = 1, \omega^{11} = 1, so that a root is
\beta = \omega + \omega^{10} \beta = \omega + \omega^{10}
\beta^2 = \omega^2 + 2 + \omega^9, \beta^2 = \omega^2 + 2 + \omega^9,
\beta^3 = \omega^3 + 3 \omega + 3 \omega^{10} + \omega^8, \beta^3 = \omega^3 + 3 \omega + 3 \omega^{10} + \omega^8,
\beta^4 = \omega^4 + 4 \omega^2 + 6 + 4 \omega^9 + \omega^7, \beta^4 = \omega^4 + 4 \omega^2 + 6 + 4 \omega^9 + \omega^7,
\beta^5 = \omega^5 + 5 \omega^3 + 10 \omega + 10 \omega^{10} + 5 \omega^8 + \omega^6. \beta^5 = \omega^5 + 5 \omega^3 + 10 \omega + 10 \omega^{10} + 5 \omega^8 + \omega^6.
Then
1 + 3 \beta – 3 \beta^2 – 4 \beta^3 + \omega^4 + \omega^5 = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + \omega^7 + \omega^8 + \omega^9 + \omega^{10} = \frac{\omega^{11} – 1}{\omega – 1} = 0 1 + 3 \beta – 3 \beta^2 – 4 \beta^3 + \omega^4 + \omega^5 = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + \omega^7 + \omega^8 + \omega^9 + \omega^{10} = \frac{\omega^{11} – 1}{\omega – 1} = 0