jmc

Factorising the numerator gives us $$\frac{3x^3-x^2-10x}{q(x)}=\frac{x(x-2)(3x+5)}{q(x)}.$$There will only be a hole at $x=2$ if both the numerator and denominator are $0$ when $x=2$. We can see that this is already true for the numerator, hence $q(x)$ must have a factor of $x-2$.

Since there is a vertical asymptote at $x=-1$, $q(-1)=0$. By the Factor theorem, $q(x)$ must have a factor of $x+1$.

Since there is no horizontal asymptote, we know that the degree of $q(x)$ must be less than the degree of the numerator. The numerator has a degree of $3$, which means that $q(x)$ has degree at most $2$.

Putting all of this together, we have that $q(x)=a(x-2)(x+1)$ for some constant $a$. Since $q(1)=-6$, we have $a(1-2)(1+1)=-6$. which we can solve to get $a=3$. Hence, $q(x)=\boxed{3(x-2)(x+1)}=3x^2-3x-6$.