does this mean the function equals 0 at 1 and -2??
Vertical asymptotes, for rational functions, occur where the denominator is 0. Perhaps that is what you intended. Further, a quadratic of the form $\displaystyle x^2+ ax+ b$ is 0 at $\displaystyle x= x_0$ and $\displaystyle x= x_1$ if and only if $\displaystyle x^2+ ax+ b= (x- x_0)(x- x_1)$.