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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.

 Feb 25, 2021
 #1
avatar+8457 
+1

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.

 Feb 25, 2021

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