$V$ is the vector space of polynomials $\mathbb R[X]_{<4}$. Let $U = \{P \in V : P(3) = 0\}$.
1) Find a basis of $V/U$.
2) Write down the matrix that represents the canonical mapping $can : V → V /U $sending $P$ to $P + U$ in terms of the basis $\{X^3,X^2,X,1\}$ of $V$ and the basis $V/U$.
I know that the basis of $U=\{X-3,X(X-3),X^2(X-3)\}$
$V/U$ is the set of cosets of $U$, which is $\{P+U | P \in V\}$ However, I'm having trouble finding the basis of it. I think it will be all those elements in $V$ minus all elements in the basis of $U$.
2) The answer is $[27,9,3,1]$ but I don't know how to achieve it.
For the basis of $V/U$ note that the class of equivalence of every vector in $U$ is the zero vector of the quotient space, so a good approach would be taking that base of $U$ you already know and extend it to a base (lets call it $B$) of $V$, then take the classes of equivalence of those vectors of $B$, they will be all zero (the zero of the quotient space) except for one, and that one generates the whole quotient space. In general (i don't know if you are allowed to use that result) the dimension of a quotient space $V/U$ is the dimension of $V$ minus the dimension of $U$.
For the matrix, do you know how you get in general the matrix associated to a linear transformation? You have to take the $i$-th vector of the basis of the domain you are considering, apply it the transformation and then the take result of that and write a vector whose coordinates are the coordinates of that result in the basis of the codomain you are considering, that vector will be the $i$-th column vector of your matrix. In this case you must take $x^3$, apply to it the canonical map and find the coordinate of it in the basis of $V/U$ you found (which is not unique, you may not arrive to the same result of your textbook), and that will be the first entry of your matrix, then do the same for the other vectors of the standard basis of $V$