jmc

The fact that the function can be simplified to a quadratic means we can probably divide $(x+2)$ out of the numerator after factoring the numerator into $(x+2)$ and the quadratic $A{x}^2+Bx+C$. Using long division or synthetic division, we find that the numerator breaks down into $(x+2)$ and $(x^2+6x+9)$.

Now, we have $$y=\frac{(x+2)(x^2+6x+9)}{x+2}.$$After we divide out the $x+2$, we're left with $x^2+6x+9$, so $A=1$, $B=6$, and $C=9$.

The domain of the quadratic function is all real numbers, but our original function was undefined when the denominator $x+2$ equaled 0. After dividing out the $x+2$ we still have to take into account that the function is undefined at $x+2=0$. So, the function is not defined at $x=-2$, giving us our value for $D$. Therefore, $A+B+C+D=1+6+9+(-2)=\boxed{14}$.