$$(x-9)/(√x - 3)$$
Note that the $3$ of the denominator is not part of the square root (just $x$).
Additionally, the hole of this expression is $(9,6)$. I don't know how to reach this result, aside from the fact that I must first find a common factor between the numerator and denominator.
I've attempted multiplying by $x$, which produces $(x^2-9x)/(x-3x)$, but this doesn't seem to help. Somebody point me in the right direction please.