Equivalence class of polynomials $X$ is the set of all polynomials over $\mathbb{R}$. We define an equivalence relation on $X$ such that $p$~$q$ iff $p(0)=q(0)$.
($1$) What is the equivalence class of $p(x)=x$?
($2$) Give a description of $X$/~ by showing a set bijective to $X$/~.

For ($1$), I think the equivalence class is something like all polynomials with zero constants, so $p(x)$ s.t. $\forall p(x)$, $p(0) = 0$.
For ($2$), I've started by defining a function $f:X \to$ $X$/~ s.t. $p(x) = x$ for $\forall x \in X$ but am pretty sure this isn't right.
 A: Based on your comments, we see that this is an exercise in set theory, not in ring theory.
Relations like yours $\rm\ p\sim q \smash[t]{\overset{\ def}{\iff}} f(p) = f(q),\: $ for $\rm\:f(p) =   p(0),\:$ are always equivalence relations.
Generally, suppose $\rm\,\ u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\ $ for some function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation properties of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f\,$ as follows:


*

*reflexive  $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$

*symmetric  $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$

*transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$
Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$ 
For $(2)$ you are correct, the equivalence class of $\rm\:p(x)=x\:$ is precisely the set of all polynomials $\rm\:q\:$ such that $\rm\:q(0) = p(0) = 0.\:$
Hint for $(3)\!:\:$ choose one "simple" representative of each equivalence class, e.g. one of least degree. 
A: Your solution for (1) is correct. For (2) an idea: define a function
$$\phi:\Bbb R[x]\to \Bbb R\;,\;\;\;\phi(p(x)):=p(0)$$ 
(i) Check $\,\phi\,$ is a surjective ring homomorphism
(ii) Show that $\,\ker\phi=\langle x\rangle\,$ = the ideal generated by the identity polynomial.
(iii) Use the first isomorphism theorem.
Added by OP's comment: A simpler (I think) approach could be to define a map of set
$$\phi: X/\sim\;\longrightarrow\Bbb R\,\;\;,\;\;\phi[f(x)]:=f(0)$$
a) Check the map above is well defined, i.e.: $\,[f(x)]=[g(x)]\Longrightarrow f(0)=g(0)\,$
b) Check the map $\,\phi\,$ is injective and surjective.
A: Hint: If $p = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$, then $p(0) = a_0$. So two polynomials are equivalent iff ...
