Tricky polynomial equation

Question

Is there any way of solving this equation for $x$ $$-1 + x+ ax(a+bx^h)^4 = 0$$ where $h\in\mathbb{N}$, $a,b\in \mathbb{R}$.

This is the most simplified version of such equation I could write, and I suspect this is not possible to solve in general, but I was wondering if something could be said about $x$. Any ideas?

The motivation behind such equation comes from a previous question I made.

I think the best that can be done is approach the roots of the polynomial. — Tiago Cavalcante, Jan 03 '22 at 13:40
An equivalent equation for $x\in\Bbb R$ is $x^{h+\frac14}+\frac ab x^\frac14=\pm\frac1{b\sqrt[4]{a+1}}$. You can substitute $x^\frac14=w$ and get a general trinomial equation which have closed forms using FoxH and another Wolfram function. — Тyma Gaidash, Dec 08 '23 at 23:14

Claude Leibovici · Answer 1 · 2022-01-12T03:00:28.677

This must be tried before concluding.

Let me suppose that you are looking for the first zero of function $$f(x)=-1 + x+ a\,x\,(a+b\,x^h)^4 $$ Expanding it and using a kind of series reversion, we have, as an estimate, the pattern $$x\sim\frac 1{1+a^5}-\frac {4a^4b}{(1+a^5)^{h+2}}+\frac {2a^3b^2\big[(8h+5)a^5-3\big]}{(1+a^5)^{2h+3}}-$$ $$\frac{4 a^2b^3\big[(24h^2+22h+5)a^{10}-2(9h+5)a^5+1 \big]}{(1+a^5)^{3h+4}}+\cdots$$

Tricky polynomial equation

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