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

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(C) We have,$\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.First,perform polynomial division: $\frac{x

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(C) We have,$\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.First,perform polynomial division: $\frac{x^3}{(2x - 1)(x^2 - 2x + 1)} = \frac{x^3}{2x^3 - 5x^2 + 4x - 1}$.Dividing $x^3$ by $2x^3 - 5x^2 + 4x - 1$,we get the quotient $A = \frac{1}{2}$ and the remainder $\frac{5x^2 - 4x + 1}{2}$.So,$\frac{x^3}{(2x - 1)(x - 1)^2} = \frac{1}{2} + \frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2}$.Now,decompose $\frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2} = \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.Multiplying by $(2x - 1)(x - 1)^2$,we get $5x^2 - 4x + 1 = 2B(x - 1)^2 + 2C(2x - 1)(x - 1) + 2D(2x - 1)$.For $x = 1$: $5 - 4 + 1 = 2D(2 - 1) \Rightarrow 2 = 2D \Rightarrow D = 1$.For $x = \frac{1}{2}$: $5(\frac{1}{4}) - 4(\frac{1}{2}) + 1 = 2B(\frac{1}{2} - 1)^2 \Rightarrow \frac{5}{4} - 2 + 1 = 2B(\frac{1}{4}) \Rightarrow \frac{1}{4} = \frac{B}{2} \Rightarrow B = \frac{1}{2}$.Comparing coefficients of $x^2$: $5 = 2B + 4C \Rightarrow 5 = 2(\frac{1}{2}) + 4C \Rightarrow 5 = 1 + 4C \Rightarrow 4C = 4 \Rightarrow C = 1$.Finally,calculate $2A - 3B + 4C + 5D = 2(\frac{1}{2}) - 3(\frac{1}{2}) + 4(1) + 5(1) = 1 - 1.5 + 4 + 5 = 8.5 = \frac{17}{2}$.Wait,re-evaluating the partial fraction decomposition: $\frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2} = \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.Using $A=1/2, B=1/2, C=1, D=1$,the expression $2A - 3B + 4C + 5D = 2(1/2) - 3(1/2) + 4(1) + 5(1) = 1 - 1.5 + 9 = 8.5$.

If $\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$,then $2A - 3B + 4C + 5D = $

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