Given frac{3x-2}{(x+1)^2(x+3)} = frac{A}{x+1} | vedclass

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(B) Given the partial fraction decomposition: $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$.Multiplying both sides b

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$-5$

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(B) Given the partial fraction decomposition: $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$.Multiplying both sides by $(x+1)^2(x+3)$,we get: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.Expanding the terms: $3x-2 = A(x^2+4x+3) + B(x+3) + C(x^2+2x+1)$.Equating coefficients of $x^2$: $A + C = 0 \Rightarrow C = -A$.Equating coefficients of $x$: $4A + B + 2C = 3$.Equating constants: $3A + 3B + C = -2$.Substitute $C = -A$ into the other equations:$4A + B - 2A = 3$ $\Rightarrow 2A + B = 3$ $\Rightarrow B = 3 - 2A$.$3A + 3(3 - 2A) - A = -2$ $\Rightarrow 3A + 9 - 6A - A = -2$ $\Rightarrow -4A = -11$ $\Rightarrow A = \frac{11}{4}$.Then $C = -\frac{11}{4}$ and $B = 3 - 2(\frac{11}{4}) = 3 - \frac{11}{2} = -\frac{5}{2}$.We need to find $4A + 2B + 4C = 4(\frac{11}{4}) + 2(-\frac{5}{2}) + 4(-\frac{11}{4}) = 11 - 5 - 11 = -5$.

Given $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$,find the value of $4A + 2B + 4C$.

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