Finding C and D is called partial fraction decomposition. See an example here and also in section Examples on the same page.

You can do that simply by adding the fractions as indicated: $\displaystyle \dfrac{C}{x- 3}+ \dfrac{D}{x+ 8}= \dfrac{C(x+8)}{(x- 3)(x+ 8)}+ \dfrac{D(x- 3)}{(x-3)(x+ 8)}$$\displaystyle = \dfrac{(C+D)x+ (8C- 3D)}{(x- 3)(x+ 8)}= \dfrac{4x- 23}{(x- 3)(x+ 8)}$
Do you see that you must have C+ D= 4 and 8C- 3D= -23?

Surely you see that $\displaystyle (x- 3)(x+ 8)= x^2+ 5x- 24$?