Let $V$ be the space of polynomials of degree less than or equal to $N$, where $N \ge 3$, and, abusing notation slightly, we allow $N = \infty$ to mean 'all polynomials'.
Suppose $p \in V$, where $p(x) = \sum_{k=0}^n p_k x^k$. Then if $w(x) = \sum_{k=4}^n p_k x^k = x^4 \sum_{k=0}^n p_k x^{k-4}$, we see that $w \in W$. Hence $x \mapsto \sum_{k=0}^3 p_k x^k \in [p] = \{ p \} +W$. Since $p$ was arbitrary, it follows that $\dim V/W \le 4$.
Let $e_k(x) = x^k$ for $k=0,..,3$. Suppose $\sum_{k=0}^3 \alpha_k [e_k] = [0]$, or equivalently, $[\sum_{k=0}^3 \alpha_k e_k] = [0]$. This implies that $\sum_{k=0}^3 \alpha_k e_k = (x \mapsto \sum_{k=0}^3 \alpha_k x_k) \in W$. However, the only way $\sum_{k=0}^3 \alpha_k e_k$ can be divisible by $x \mapsto x^4$ is if $\alpha_k = 0$. Hence $\{[e_k] \}_{k=0}^3$ are linearly independent, and hence we have $\dim V/W = 4$.
If $N<3$, a slight modification of the above shows that $\dim V/W = \min(4,N+1)$.
I know this is an old question, but it seems to me that no one has answered it in a "correct" fashion yet, concerning "infinite" of course.
The most generalized definition I've seen is the next one:
Let H be a subspace in a vector space X. H is called hyperplane if $H \neq X$ and for every subspace V such that $ H \subseteq V $ only one of the following is satisfied: $ V = X$ or $ V = H $.
Lemma: A subspace H in X is hyperplane iff e is in X \ H such that $$ <\{e,H \}> = \{ \lambda e + h : \lambda \epsilon \mathbb{R}, h \epsilon H\} = X $$
Just for fun:
Theorem: A subspace H in a vector space X is hyperplane iff there is a non-zero linear function $$ l : X \to \mathbb{R}$$ such that $$ H = \{x \epsilon X : l(x) = 0\} = Ker( l ) $$
Good luck proving this.
Best Answer
Define the map $T\colon V\rightarrow \mathbb{R}^4$ by $$T\left(\sum_{i=0}^n a_ix^i\right) = (a_0,a_1,a_2,a_3),$$ meaning that you keep only the first four coefficients of any polynomial. This is a surjective linear map, and its kernel is $$\text{ker}(T) = \left\{\sum_{i=4}^n a_ix^i\ |\ n\in\mathbb{N}, a_i\in\mathbb{R}\right\}.$$ Note that $\text{ker}(T) = W$. The induced map $$\tilde{T}\colon V/W\rightarrow\mathbb{R}^4$$ is thus an isomorphism.