Find a linear transformation $T:P_2(\mathbb R)\to\mathbb R$, $P_2(\mathbb R)$ being the vector space of polynomials of degree no more than $2$, such that $\ker T=U$ where $U=\{P(x)\in P_2(\mathbb R)\mid P(4)=0\}$.
A vector $p\in P_2(\mathbb R)$ could be written as $p=ax^2+bx+c$, so any vector $u\in U$ could then be expressed as
$$u=ax^2+bx+(-16a-4b)=a(x^2-16)+b(x-4)$$
Then $u\in\mathrm{span}\{x^2-16,x-4\}$. Where do I go from here to find a possible $T$?
I know that both of these vectors are linearly independent. I've seen similar questions/examples that suggest I "extend this set of vectors to a basis of insert vector space". Here, I imagine it means I need to add another vector to the spanning set so that it forms a basis of $P_2(\mathbb R)$, which I believe would be satisfied by $1$ because no linear combination of the vectors in the spanning set can yield a constant.
The aforementioned examples went to say that any possible transformation $T$ should satisfy
$$\begin{cases}T(x^2-16)=0\\T(x-4)=0\\T(1)\neq0\end{cases}$$
and I see why this should be the case, but I don't know what $T$ is supposed to look like, as I'm only really used to working with transformations $T:\mathbb R^m\to\mathbb R^n$. Would it necessarily be a matrix? and of what dimensions?
More seriously, your base for $\ker(T)$ could be simplified, but whatever.
Complete this family into a base of $\mathbb R_2[X]$, by adding the constant polynomial $1$.
You then just have to define $T$ by stating : $T(1)=1$, $T(X-4)=0$ and $T(X^2-16)=0$.
As for every polynomial $P=aX^2+bX+c$, you have $$P=a(X^2-16)+b(X-4)+c+16a+4b$$ you can check that $T(P)=P(4)$. Bingo !