I recently saw this question about polynomials that almost satify the Eisenstein's criterion.
Let $f(x) = \sum_{i=0}^n a_nx^n\in \mathbb{Z}[x]$. Suppose there is a prime $p$ such that $p$ does not divide $a_{n-1}$ and $a_n$, $p$ divides the other coefficients and $p^2$ does not divide $a_0$, then $f$ is irreducible over $\mathbb{Q[x]}$ if and only if $f$ has no rational root.
From the answer, it was shown that if $f$ can be reduced, say $f=gh$, then either $g$ or $h$ is linear.
Now, here are my questions.
Am I correct that the approach in the answer is applicable to UFDs? Of course, we replace $\mathbb{Q}$ by the field of fractions. I can't see any special property of $\mathbb{Z}$ used that is not satisfied by any UFD.
Is my interpretation of the proof correct that at the "worst" case, $f$ can only be reduced to a product of a linear polynomial and a polynomial of degree $n-1$? Specifically, if $f=gh$, then $g$ and $h$ are already the irreducible factors of $f$?