Describe the elements of the quotient space of $V/S$ and also find a basis for it.
$V = P_{10}$, the space of all real polynomials of degree at most 10, $$S=\operatorname{span}(\{{x,x^3,x^5,x^7,x^9}\}) .$$
Attempt: To my understanding of a quotient space $V/S$ means $V$ mod $S$. The elements of the quotient space is the set of even degree polynomials with degree of at most $10$ since taking the modulo of the set would essentially only put things that were not divided out(remainder) and a basis would be $\{x^2,x^4,x^6,x^8,x^{10}\}$
I'm unsure what this quotient space means in a practical sense and my textbook is very abstract. So if someone could tell me where I'm thinking about it wrong that would help immensely. My textbook mentions that a quotient space is the set of equivalence classes. But I'm unsure looking at the definition what that means for the scope of this sort of question.
In the following I will use your notation and I will assume $V$ is a $K$-vector space with $K$ field.
By definition, an element of $V/S$ is an equivalence class of the following relation: \begin{equation} v\sim w \Longleftrightarrow v-w\in S \end{equation} For example the element $x^2$ and $x^2+5x^7$ are in the same class (the difference is $5x^7\in S$) so they are the same element in $V/S$.
A description of $V/S$ is the following: \begin{equation} V/S = \{a_0 + a_2x^2 + ... + a_{10}x^{10} +S \ | \ a_i\in K\} \end{equation} and with this notation I mean \begin{equation} a_0 + a_2x^2 + ... + a_{10}x^{10} +S = \{a_0 + a_2x^2 + ... + a_{10}x^{10} +S \ | \ s\in S\} \end{equation} In order to prove that they are the equivalence's classes, we have to verify two fact:
Every element of $V$ lies in one of these classes: this is easy because taking $v=b_0+b_1x+b_2x^2+...+b_{10}x^{10}$ we have \begin{equation} v = (b_0+b_2x^2+...+b_{10}x^{10}) + (b_1x^1+ ... b_9x^9) \in b_0+b_2x^2+...+b_{10}x^{10} +S \end{equation}
Every element of $V$ lies in exactly one class: suppose $v$ lies in two different classes we have: \begin{equation} (a_0+a_2x^2+...+a_{10}x^{10}) + s_1 = v = (b_0+b_2x^2+...+b_{10}x^{10}) + s_2 \end{equation} Taking the difference we have \begin{equation} (a_0-b_0)+(a_2-b_2)x^2+...+(a_{10}-b_{10})x^{10} = s_2-s_1 \in S \end{equation} and this implies $a_i=b_i$ for $i=\{0,2,4,6,8,10\}$. So the two classes are the same.
The discussion above provides a description of $V/S$ as a set. But we have also a natural structure of vector space induced by the structure of $V$. This means we can do linear combinations of elements of $V/S$ in this way: \begin{gather} \alpha [v_1] := [\alpha v_1] \\ [v_1]+ [v_2] := [v_1 + v_2] \end{gather} where $\alpha \in K$ and $[v_1],[v_2]$ are the equivalence's classes in $V/S$ of $v_1$ and $v_2$. These operations are well defined.
In our example, these operations correnspond to: \begin{gather} \alpha (a_0 + a_2x^2 + ... + a_{10}x^{10} +S) = \alpha a_0 + \alpha a_2x^2 + ... + \alpha a_{10}x^{10} +S\\ (a_0 + ... + a_{10}x^{10} +S) + (b_0 + ... + b_{10}x^{10} +S) = (a_0+b_0) + ... + (a_{10}+b_{10})x^{10} +S \end{gather}
Now it's really easy to see that a base of $V/S$ is \begin{equation} \mathcal B = \{1+S, x^2 + S , x^4 +S , x^6+S, x^8+S, x^{10}+S\}\subset V/S \end{equation}