If frac{27x^2+32x+16}{(3x+2)^2(1-x)} = frac{A | vedclass

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(B) Given the partial fraction decomposition: $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$.Multiplying b

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$12$

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(B) Given the partial fraction decomposition: $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$.Multiplying both sides by $(3x+2)^2(1-x)$,we get:$27x^2+32x+16 = A(3x+2)(1-x) + B(1-x) + C(3x+2)^2$.Setting $x = 1$: $27(1)^2 + 32(1) + 16 = C(3(1)+2)^2$ $\Rightarrow 75 = 25C$ $\Rightarrow C = 3$.Setting $x = -2/3$: $27(-2/3)^2 + 32(-2/3) + 16 = B(1 - (-2/3))$ $\Rightarrow 12 - 64/3 + 16 = B(5/3)$ $\Rightarrow 20/3 = 5B/3$ $\Rightarrow B = 4$.Setting $x = 0$: $16 = A(2)(1) + B(1) + C(2)^2$ $\Rightarrow 16 = 2A + 4 + 4(3)$ $\Rightarrow 16 = 2A + 16$ $\Rightarrow A = 0$.Finally,$AB + BC + CA = (0)(4) + (4)(3) + (3)(0) = 0 + 12 + 0 = 12$.

If $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$,then $AB+BC+CA =$

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