Euclidean division in quotients of polynomial rings

I know that given a field $$\mathbb{K}$$, the one variable polynomial ring $$\mathbb{K}[x]$$ is an euclidean domain. This helps to figure out how the quotient $$\dfrac{\mathbb{K}[x]}{(f(x))}$$ (where $$f(x) \in \mathbb{K}[x]$$) is made: its elements are the polynomials $$h(x)$$ such that $$\text{deg}(h(x))<\text{deg}(f(x))$$, because of the euclidean division. If we have a generic $$A[x]$$ where $$A$$ is a commutative ring with $$1$$, what can we say about $$\dfrac{A[x]}{(f(x))}$$ with $$f(x) \in A[x]$$?

Let $$c \in A$$ be the leading coefficient of $$f$$. If $$c$$ is a unit, then every element of $$A[x]/(f)$$ can still be represented as a polynomial in $$x$$ of degree less than $$\deg(f)$$, by exactly the same argument. However, if $$c$$ isn't a unit, the situation is a bit more complicated.
For example, consider $$\newcommand{\ZZ}{\mathbb{Z}}\ZZ[x]/(2x - 1)$$. If $$g \in \ZZ[x]$$ has odd leading coefficient, then for any $$h \in \ZZ[x]$$, we have $$\deg(g + (2x - 1)h) \geq \deg(g)$$. In particular, any system of representatives of $$\ZZ[x]/(2x - 1)$$ requires polynomials of arbitrarily high degree. (One can also see this by observing that $$\ZZ[x]/(2x - 1) \cong \ZZ[\frac{1}{2}]$$, and one can have arbitrarily high powers of $$2$$ in the denominator of elements of $$\ZZ[\frac{1}{2}]$$.)