If frac{9}{(x - 1)(x + 2)^2} = frac{A}{x - 1} | vedclass

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Question

(C) Given the partial fraction decomposition: $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$.Multiplying both

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$5$

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(C) Given the partial fraction decomposition: $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$.Multiplying both sides by $(x - 1)(x + 2)^2$,we get: $9 = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1)$.For $x = 1$: $9 = A(1 + 2)^2$ $\Rightarrow 9 = 9A$ $\Rightarrow A = 1$.For $x = -2$: $9 = C(-2 - 1)$ $\Rightarrow 9 = -3C$ $\Rightarrow C = -3$.Equating the coefficients of $x^2$ on both sides: $0 = A + B \Rightarrow B = -A = -1$.Therefore,$A - B - C = 1 - (-1) - (-3) = 1 + 1 + 3 = 5$.

If $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$,then $A - B - C = $

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