jmc

Expanding $(x+2)(3x^2-x+5)$ gives $$\begin{array}{rl}& x(3x^2)+x(-x)+x(5)+2(3x^2)+2(-x)+2(5)\\ & =A{x}^3+B{x}^2+Cx+D.\end{array}$$Computing the products on the left side gives $$3x^3-x^2+5x+6x^2-2x+10=A{x}^3+B{x}^2+Cx+D.$$Simplifying the left side gives $$3x^3+5x^2+3x+10=A{x}^3+B{x}^2+Cx+D,$$so $A=3$, $B=5$, $C=3$, and $D=10$ and $$A+B+C+D=3+5+3+10=\boxed{21}.$$