jmc

Since the given function has vertical asymptotes at $2$ and $-2$, we know that $q(2)=q(-2)=0$ (i.e. $2$ and $-2$ are roots of $q(x)$). Moreover, since the given function has no horizontal asymptote, we know that the degree of $q(x)$ must be less than the degree of the numerator, which is $3$.

Hence, $q(x)$ is a quadratic with roots $2$ and $-2$. In other words, we can write it as $q(x)=a(x+2)(x-2)$ for some constant $a$. Since $q(3)=15$, we have $a(3+2)(3-2)=15$.

Solving for $a$ gives $a=15/5=3$. Hence $q(x)=3(x-2)(x+2)=\boxed{3x^2-12}$.