Find the product CD of the integers C and D for which
\[\frac{C}{x-3}+\frac{D}{x+8}=\frac{4x-25}{x^2+5x-24}\]
for all real values of x except -8 and 3.
+2666 Multiplying the entire equation by \(x^2 + 5x - 24 \) gives us: \((x + 8)C + (x-3)D = 4x - 25\). Note that \(x^2 + 5x - 24 = (x - 3) ( x+8)\)
Expanding the left-hand side gives us: \(Cx + 8C + Dx - 3D = 4x - 25\).
Note that the only 2 terms with x in them must sum to 4x, so we have the equation: \(Cx + Dx = 4x \), meaning \(C + D = 4\)
Also note that the only 2 constants must sum to -25, giving us the equation: \(8C - 3D = -25\).
Now, we have the following system:
\(8C - 3D = -25\)
\(C + D = 4\)
However, solving this system yields \(C = -{13 \over 11}\) and \(D = {57 \over 11}\).
Thus... there are no integer solutions, and no product as well....
Here is the WA link: https://www.wolframalpha.com/input?i2d=true&i=Divide%5BC%2Cx+-+3%5D+%2B+Divide%5BD%2Cx+%2B+8%5D+%3D+Divide%5B4x+-+25%2CPower%5Bx%2C2%5D%2B5x+-+24%5D
