jmc

Let $r$ and $s$ be the roots of $a{x}^2+bx+c=a(x-r)(x-s),$ so $$\frac{1}{a(x-r)(x-s)}=\frac{A}{x-r}+\frac{B}{x-s}.$$Multiplying both sides by $(x-r)(x-s),$ we get $$A(x-s)+B(x-r)=\frac{1}{a}.$$Expanding, we get $$(A+B)x-As-Br=\frac{1}{a}.$$Since this equation represents an identity, the coefficients of $x$ on each side must match. In other words, $A+B=\boxed{0}.$