+206 Find the product CD of the integers C and D for which
\[\frac{C}{x-3}+\frac{D}{x+8}=\frac{4x-23}{x^2+5x-24}\]
for all real values of x except -8 and 3.
+9674 Multiply through by \(x^2 + 5x - 24\).
For all real values of x except -8 and 3,
\(C(x + 8) + D(x - 3) = 4x - 23\\ (C + D)x + (8C - 3D) = 4x - 23\)
Comparing coefficients,
\(\begin{cases}C + D = 4\\8C - 3D = -23\end{cases}\)
Solving, (C, D) = (-1, 5).
Therefore CD = -5.
