smc

$a_0$ cannot be $0$, because $a_2\ne 0$, so the polynomial can take non-zero values when $a_0=0$ and thereby not satisfy the equation. Expanding $a_0(1-\frac{x}{r})(1-\frac{x}{s})$ gives us $a_0(1-\frac{x}{r}-\frac{x}{s}+\frac{x^2}{rs})$. Using Vieta's formulas, we see that $rs=\frac{a_0}{a_2}$, so $a_0=a_{2}rs$. Plugging this into the expanded right hand side of the given equation, we see that that side equals $a_0-a_2(r+s)x+a_2{x}^2$. Using Vieta's formulas again, we equate this expression to $a_0+a_{1}x+a_2{x}^2$, which is the left hand side of the given equation. Thus, as long as $a_0\ne 0$, the equation holds for all values of $x$. This fact corresponds with answer choice $\boxed{\text{for all values of }x,a_0\ne 0}$.