Let $I = \{ f(x) \in \mathbb{C}[x] :f(1)=f'(1)=0\}$. One can show that $I$ is an ideal in $\mathbb{C}[x]$. Thus, $\mathbb{C}[x]/I$ is a ring. Viewed as a vector space over $\mathbb{C}$, what is the dimension of $\mathbb{C}[x]/I$?
Let $f(x) \in I$, we must have $f(x) = (x-1)g(x)=xg(x) - g(x)$ for some $g(x)\mathbb{C}[x]$. In addition, we also have $g(1)=0$ since we need $f'(1) = 0$. How do I find the dimension from here? I am thinking about the possible remainder when I divide some polynomial $p(x)$ by $f(x)$, but I don't know the degree of $f(x)$.
You can show that a root $c \in F$ of a polynomial $f(x) \in F[x]$ is a repeated root of $f$ $ \iff c$ is also a root of $f'(x)$. Thus, $I$ is the set of polynomials with $1$ as a repeated root. Since $\mathbb{C}[x]$ is a principle ideal domain, it must be the case that $I = \langle (x-1)^2 \rangle$.
Now think about the elements of $\mathbb{C}[x]/\langle (x-1)^2 \rangle$. The equivalence classes of this quotient ring will be the possible remainders when polynomials in $\mathbb{C}[x]$ are divided by $(x-1)^2$...