Possible misconception of the ring of polynomials over a field F. When we let $F$ be a field and we define $R = F[x]$ to be the ring of polynomials where each coefficient is in $F$, how exactly can we talk about divisibility of polynomials? Do we ignore the actual value of the polynomial, and only consider the variable as 'place-holder' for the coefficients? If not, then what do we do about the roots of the denominators?
I guess what I'm trying to get at is whether there's a difference between f(x), the value the polynomial f has at x, and the polynomial f. 
 A: In ring theory, one must distinguish between an (abstract) polynomial and a polynomial function. An (abstract) polynomial in $p \in F[x]$ is (formally) just a map $f : \Bbb N_{0} \to F$ with finite support (all but finitely many of the outputs of $p$ are $0$). Addition of polynomials is standard component-wise addition of functions and multiplication is given by
$$
(p\cdot q)_n = \sum_{k=0}^n p_kq_{n-k}.
$$
Here, $p_n$ is the $n$th coefficient of the polynomial $p$. Associated with every polynomial and $\alpha \in F$, there is an evalutation homomorphism $\text{eval}_\alpha : F[x] \to F$ defined by 
$$
\text{eval}_{\alpha}(p)=\sum_{n=0}^{\deg p} p_n \alpha^n.
$$
($\deg p$ is defined to be the largest $k$ such that $p_k \ne 0$.) A function $f : F \to F$ is called a polynomial function if there exists an (abstract) polynomial $p \in F[x]$ such that $f(\alpha) = \text{eval}_\alpha(p)$ for all $\alpha \in F$.
The distinction between polynomial and polynomial function is critical when one moves to finite fields. Consider the (abstract) polynomial $p(x) = 1 + 1\cdot x$ and the polynomial function $f(x) = 1 + x$ for the field $\Bbb Z / 2\Bbb Z$. Then $f(0) = f(1) = 0$ so the function $f$ is equal to the zero function but $p$ is not the zero polynomial. 
To divisibilty! An (abstract) polynomial $p$ divides a polynomial $g$ if there exists another polynomial $q$ such that $g = p \cdot q$. We don't care if $p$ has zeros. Heck, $p$ could be the example above that is entirely zero as a polynomial function. All we care about is that $p$ is not the zero polynomial.

Edit to Address the OP's questions in the comments: Above I mean $\Bbb N_0 := \{0,1,2,3,4,\ldots\}$--different mathematicians define $\Bbb N$ as either $\{1,2,3,\ldots\}$ or $\{0,1,2,3,\ldots\}$ so I subscript $0$ to indicate I am using the latter definition. By finite support, I mean that, for an (abstract) polynomial $p$, the sequence $p_0,p_1,p_2,\ldots$ is eventually all zeros. Formally, there exists an $N \in \Bbb N_0$ such that $p_{m} = 0$ for every $m > N$. The finite support condition is necessary to rule out infinite series like $p(x) = 1 + x + x^2 + x^3 + \cdots$. You're absolutely right that an abstract polynomial has no output: it's just a formal sequence of coefficients.
The evaluation homomorphism is the "bridge" between an abstract polynomial and its corresponding polynomial function. If I have the abstract polynomial $p(x) = 1 + x + 2x^2$, the evaluation homomorphism allows me to actually evaluate this polynomial as you would have in high school algebra. (e.g. $\text{eval}_2(p) = 1 + 2 +2\cdot 2^2 = 11$.)
