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 $3$.
I think the first step here is to find the least common denominator, which I believe is $(x-3)(x+8)$. Yes, now we need to multiply both sides of the equation by $(x-3)(x+8)$. So the left side of the equation becomes $C(x+8) + D(x-3)$. Which becomes $Cx + 8C + Dx - 3D$. That expression is equal to $(C+D)x + (8C-3D)$. Now let's focus on the right side of the equation. Right, we need to FOIL out $(x-3)(x+8)$. That becomes $x^2 + 5x - 24$. And we're given that the right side of the equation is equal to $\frac{4x-23}{x^2+5x-24}$. So we have the equation $(C+D)x + (8C-3D) = \frac{4x-23}{x^2 + 5x - 24} \cdot (x^2 + 5x - 24)$. The right side of that equation becomes $4x-23$. So we have $(C+D)x + (8C-3D) = 4x - 23$. Looking at the coefficients of the $x$ terms, we get $C+D = 4$. And looking at the constant terms, we get $8C - 3D = -23$. So we have the system of linear equations $C+D = 4$ and $8C - 3D = -23$. Multiplying the first equation by $8$ gives $8C+8D=32$. Subtracting the second equation from that gives $11D=55$. So $D=5$. That means $C=4-D=4-5=-1$. So the product $CD$ is $(-1)(5)=-5$.

