Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$. The obtained value, $m$, is an exact divisor of: 

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$
Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$, where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$. The only number that fits the first equation is $y=6$, so $m=-18$. The only choice that is a multiple of 18 is $\boxed{\textbf{(C) }36}$.