How to construct a degree $4$ polynomial $h(z)$ such that $h(z)$ has a triple root and $h(z) - 1$ has a double root? This is a homework question, so please do not give me the full answer. I only need a hint that pushes me in the right direction.
I have been asked to construct a holomorphic function $h(z)$ whose dessin d'enfant looks like this:

For simplicity, I have assumed that $h(z)$ is a polynomial. Therefore $h$ must have degree $4$, because the preimage of a generic $t \in \mathbb C$, exemplified by those $0 < t < 1$ drawn in red, has four points. The leftmost white point is a triple root of $h(z)$, whereas the rightmost black point is a double root of $h(z) - 1$. It seems reasonable to assume that the other two black points are complex conjugates.
Taking the preceding considerations into account, I postulate that $h(z)$ is such that $h(z) = k (z-a)^3 (z-b)$ and $h(z) - 1 = k (z^2 + c^2) (z - d)^2$, for some $k < 0$. How could I construct a polynomial $h(z)$ with these characteristics?
EDIT: Never mind, I found $h(z) = 16 z^3 (2 - 3z)$. Obviously it has a triple zero at $z = 0$ and a simple zero at $z = 3/2$. Then $h(z) - 1 = - (1 - 2z)^2 (1 + 4z + 12z^2)$, which has a double zero at $z = 1/2$ and two simple zeros at two complex conjugate points.
 A: Since you've already figured out the answer, let me just give a nice reference for the study of dessins d'enfants that are trees.
Also, note that only in the case where the dessin is a tree can we assume that the associated Belyi map is a polynomial. (These are called Shabat polynomials.) When the dessin is a tree, there is only one point above $\infty$ which thus must be totally ramified. By assuming that the map is a polynomial, we are insisting that this point is $\infty$, so $\infty \mapsto \infty$. (For dessins that are not acyclic, for instance the one found here, we can't assume the associated map is a polynomial.)
Bétréma, Péré, and Zvonkin have computed a catalogue of all Belyi maps corresponding to plane trees with at most 8 edges, which is available here. On this page they give the answer $z^3(1-z)$, which when normalized so that the point with ramification index $2$ lies above $1$, yields $h(z) = \frac{256}{27}z^3(1-z)$.
We could also normalize as in the other answer. As I pointed out in this comment, the polynomial $x^4 - b x^3 - 1$ has discriminant $-27 b^4 - 256$. Taking $b$ to be a root of the discriminant, so $b = \sqrt[4]{-256/27}$, then $h(z) = z^3(z-b)$ would also work.
A: I don't think there is such a polynomial. You can assume $k=1$ and $a=0$. Then asking Wolfram Alpha to solve
$$
x^4 -bx^3 -1
$$
produces complicated expressions that look all distinct. You could probably prove they are.
https://www.wolframalpha.com/input/?i=solve+x%5E4+-ax%5E3+-1
That suggests you started down a dead end path in your search for a holomorphic function.
