Deducing Hensel's lemma for polynomials

Question

Can one somehow deduce the following theorem:

Hensel's lemma, second version: let $f(x) \in \mathbb Z_p[x]$ be a polynomial and assume that there exist polynomials $g_1(x), h_1(x) \in \mathbb Z_p[x]$ such that $g_1$ is monic, $g_1$ and $h_1$ are coprime modulo $p$ and finally $f(x) \equiv g_1(g)h_1(x) \mod p$. Then there exist polynomials $g(x), h(x) \in \mathbb Z_p[x]$ such that $g(x)$ is monic, $g(x) \equiv g_1(x)$ and $h(x) \equiv h_1(x) \mod p$ and $f(x) = g(x) h(x)$.

from the classical version:

Hensel's lemma: let $K$ be a complete ultrametric field and let $f(x) \in \mathcal O[x]$. Let $a_0 \in \mathcal O$ satisfy $|f(a_0)| < |f'(a_0)|^2$, where $f'(x)$ is the formal derivative. Then there is an $a \in \mathcal O$ such that $f(a) = 0$. Moreover, only one solution of $f(a) = 0$ satisfies $$|a-a_0| \le \frac{|f(a_0)|}{|f'(a_0)|}?$$

Answer 1

I don’t believe that there is such a proof, though I would be overjoyed to be shown one that is natural and direct.

Consider the $\Bbb Z_p$-polynomial $f(x)=px^4+(1+p^2)x^2+p$, which you can see has no roots in $\Bbb Q_p$. But you can write $f\equiv x^2\cdot1\pmod p$, so that Strong Hensel (your “second version”) applies, while there can be no direct application of Weak Hensel (your “classical version”).