On solvable octic trinomials like $x^8-5x-5=0$

Question

Solvable quintic trinomials $$x^5+ax+b=0$$ have been completely parameterized. Finding $6$th-deg versions is relatively easy to do such as, $$x^6+3x+3=0$$ which factors over $\sqrt{-3}$. No $7$th-deg are known, but surprisingly there are octic ones, such as the simple, $$x^8-5x-5=0$$ which factors over $\sqrt{5}$. And the not-so-simple ones,

$$x^8-11(4x+3)=0\\x^8+16(4x+7)=0\\x^8 + 5\cdot23^2(12 x+43) =0$$

which factors over a quartic extension (and needs the cube root of unity).

Q: Any other octic examples, if possible parametric?

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$\color{green}{Update:}$

Klajok in his answer below has found a family for the class of octic trinomials that factor over a quadratic extension. However, another class needs a quartic extension. For example, $$x^8-44x-33=0\tag1$$ which factors into four quadratics, $$x^2 + v x - (2v^3 - 7v^2 + 5v + 33)/13=0$$ and where $v$ is any root of $v^4 + 22v + 22=0$. More generally, eliminating $v$ between $$x^2 + v x + (pv^3 +qv^2 + rv + s)=0$$ $$v^4+av^2+bv+c=0$$ easily done by the resultant function of Mathematica will result in an irreducible but solvable octic and judicious choice of rational coefficients will yield a trinomial. However, it is not known if this second class of trinomials like $(1)$ has a parametric family as well.

Answer 1

Using brute force approach I have found few others, such as $$x^8+9x+9$$ $$x^8+75x+150$$

which factors over $\sqrt{-3}$ and $\sqrt{-15}$, respectively.

See https://sites.google.com/site/klajok/polynomials/x8-ax-b0

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$\color{green}{Added:}$

Given pairs of rational numbers $\left(\alpha,\beta\right)$ such that $2\alpha^2+6\alpha+1=\beta^2$. Define the following parameters:

$$u=\frac{2\alpha+1-\beta}{4},\quad v=\frac{1-\beta}{8},\quad w=\frac{\alpha}{8}\left(3\alpha-2\beta+3\right)$$ $$A=\frac{\alpha u}{2}\left(\alpha+1-4u\right),\quad B=w^2 - \alpha v^2$$

then the following identity is satisfied: $$x^8+Ax+B =\\ \left(x^4+\sqrt{\alpha}x^3+\left(-\frac{\sqrt{\alpha}}{2}+\frac{\alpha}{2}\right)x^2+\left(u\sqrt{\alpha}-\frac{\alpha}{2}\right)x+\left(v\sqrt{\alpha}+w\right)\right)\\ \left(x^4-\sqrt{\alpha}x^3+\left(\frac{\sqrt{\alpha}}{2}+\frac{\alpha}{2}\right)x^2+\left(-u\sqrt{\alpha}-\frac{\alpha}{2}\right)x+\left(-v\sqrt{\alpha}+w\right)\right)$$

Let exclude the pairs $\left(\alpha, \beta \right) \in \left\{ \left(0, -1\right), \left(0, 1\right), \left(1, 3\right) \right\}$ for which trinomials degenerate to the simpler form where $AB=0$.

Observations: If $\sqrt{\alpha}$ is not a rational number then the corresponding octic trinomials are irreducible and solvable. Otherwise the trinomials are still solvable but they are not irreducible.

Notes:

1. All the $\left(\alpha, \beta \right)$ pairs can be easily enumerated: $$\left(\alpha, \beta \right) \in \left\{ \left( \frac{2q+6}{q^2-2}, \frac{q^2+6q+2}{q^2-2} \right) : q \in \mathbb Q \setminus \lbrace -3, 4 \rbrace \right\}$$ Excluded values $q=-3$ and $q=4$ correspond to the degenerate cases $(0, -1)$ and $(1, 3)$, respectively. The remaining degenerate case $(0,1)$ corresponds to the value $q$ at infinity (I found this "subparameterization" using the following Philip Gibbs' answer: https://math.stackexchange.com/q/1905148).

2. Examples and preliminary conjectures related to this parameterization is available under the same page: https://sites.google.com/site/klajok/polynomials/x8-ax-b0 .

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I searched for more examples of $C_2 \wr A_4$ and $C_2 \wr S_4$ for $x^8+ax+b$ and the only ones for integer $|a|,|b| \leq 100000000$ are:

$C_2 \wr A_4$: $$x^8+64x+112=x^8+16(4x+7)$$

$C_2 \wr S_4$: $$x^8-44x-33=x^8-11(4x+3)$$ $$x^8+768x+1344=x^8+3\cdot8^2(4x+7)$$ $$x^8+31740x+113735=x^8+5\cdot23^2(12x+43)$$ $$x^8-856251x-2023866=x^8-3^4\cdot31^2(11x+26)$$ $$x^8-5992704x-304129728=x^8-6^4\cdot34^2(4x+203)$$

Answer 2

A result of Harris [1] is that every monic palindromic polynomial of degree-8 can be factored into two monic palindromic polynomials of degree-4.

$$ \begin{align} f(x) & = x^8 + ax^7 + bx^6 + cx^5 + dx^4 + cx^3 + bx^2 + ax + 1 \\ & = (x^4 + px^3 + qx^2 + px + 1)(x^4 + rx^3 + sx^2 + rx + 1) \\ & = x^8+x^7 (p+r)+x^6 (pr+q+s)+x^5 (p(s+1)+qr+r)+x^4 (2pr+qs+2)+x^3 (p(s+1)+qr+r)+ x^2 (pr+q+s)+x (p+r)+1 \end{align} $$

Equating the coefficients:

$$ \begin{align} a &= p+r \\ b &= pr + q + s \\ c &= p(s+1) + qr + r \\ d &= 2pr + qs + 2 \end{align} $$

For the subset of degree-8 monic palindromic polynomials of the form

$$ \begin{align} f(x) & = x^8 + 0x^7 + 0x^6 + 0x^5 + dx^4 + 0x^3 + 0x^2 + 0x + 1 \\ & = x^8 + dx^4 + 1 \\ \text{we have } 0 &= p+r \\ 0 &= pr + q + s \\ 0 &= p(s+1) + qr + r \\ d &= 2pr + qs + 2 \end{align} $$

We have the parametric solutions:

$$ r = 0 ∧ s = -q ∧ p = 0 ∧ d = 2 - q^2, \\ r = \sqrt{2} \sqrt{q} ∧ s = q ∧ p = -\sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2, \\ r = -\sqrt{2} \sqrt{q} ∧ s = q ∧ p = \sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2. \\ $$

Note that all of the above are parametrized on $d$.

On similar lines, we can factor the monic quartic palindromic polynomials into two monic quadratic palindromic polynomials each and then use the quadratic formula to get the roots.

References

[1]: J. R. Harris, "96.31 Palindromic Polynomials," The Mathematical Gazette, vol. 96, no. 536, p. 266–69, 2012. https://doi.org/10.1017/S0025557200004526