jmc

First, we factor the denominator in the right-hand side, to get $$\frac{C}{x-3}+\frac{D}{x+8}=\frac{4x-23}{(x-3)(x+8)}.$$We then multiply both sides by $(x-3)(x+8)$, to get $$C(x+8)+D(x-3)=4x-23.$$We can solve for $C$ and $D$ by substituting suitable values of $x$. For example, setting $x=3$, we get $11C=-11$, so $C=-1$. Setting $x=-8$, we get $-11D=-55$, so $D=5$. (This may not seem legitimate, because we are told that the given equation holds for all $x$ except $-8$ and $3.$ This tells us that the equation $C(x+8)+D(x-3)=4x-23$ holds for all $x$, except possibly $-8$ and 3. However, both sides of this equation are polynomials, and if two polynomials are equal for an infinite number of values of $x$, then the two polynomials are equal for all values of $x$. Hence, we can substitute any value we wish to into this equation.)

Therefore, $CD=(-1)\cdot 5=\boxed{-5}$.