TITLE: On solvable octic trinomials like $x^8-5x-5=0$
QUESTION [14 upvotes]: Solvable quintic trinomials
$$x^5+ax+b=0$$
have been completely parameterized. Finding $6$th-deg versions is relatively easy to do such as,
$$x^6+3x+3=0$$
No $7$th-deg are known, but surprisingly there are octic ones, such as the simple,
$$x^8-5x-5=0\\x^8+\tfrac12x+\tfrac7{16}=0\\x^8-44x-33=0$$
and the not-so-simple,
$$x^8 + 23^2(60 x+215) =0$$

Q: Any other octic examples, if possible parametric?


$\color{green}{Update:}$
Klajok has found a family for the class of octic trinomials that factor over a quadratic extension. However, another class needs a quartic extension, such as
$$x^8-44x-33=0\tag1$$
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.

REPLY [9 votes]: Using brute force approach I have found few others, such as 
$$x^8+9x+9$$
$$x^8+64x+112$$
$$x^8+75x+150$$
See https://sites.google.com/site/klajok/polynomials/x8-plus-ax-plus-b-is-0

$\color{green}{Added:}$
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:

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).
Examples and preliminary conjectures related to this parameterization is available under the same page: https://sites.google.com/site/klajok/polynomials/x8-plus-ax-plus-b-is-0 .


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: 
$C_2 \wr A_4$:
$$x^8+64x+112$$
$C_2 \wr S_4$:
$$x^8+44x-33$$
$$x^8+768x+1344$$
$$x^8+31740x+113735$$
$$x^8+856251x-2023866$$
$$x^8+5992704x-304129728$$