If frac{x^4}{(x-1)^2(x+1)}=Ax+B+frac{P}{(x-1) | vedclass

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(B) Given $\frac{x^4}{(x-1)^2(x+1)}$. Perform polynomial division: $\frac{x^4}{x^3-x^2-x+1} = x+1 + \frac{2x^2-x-1}{(x-1)^2(x+1)}$.Since $2x^2-x-1 = (

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$\frac{13}{4}$

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(B) Given $\frac{x^4}{(x-1)^2(x+1)}$. Perform polynomial division: $\frac{x^4}{x^3-x^2-x+1} = x+1 + \frac{2x^2-x-1}{(x-1)^2(x+1)}$.Since $2x^2-x-1 = (2x+1)(x-1)$,the expression becomes $x+1 + \frac{2x+1}{(x-1)(x+1)}$.Using partial fractions for $\frac{2x+1}{(x-1)(x+1)} = \frac{A_1}{x-1} + \frac{A_2}{x+1}$,we get $A_1 = \frac{3}{2}$ and $A_2 = \frac{1}{2}$.Thus,$\frac{x^4}{(x-1)^2(x+1)} = x+1 + \frac{3/2}{x-1} + \frac{1/2}{x+1}$.Comparing with $Ax+B+\frac{P}{x-1}+\frac{Q}{(x-1)^2}+\frac{R}{x+1}$,we get $A=1, B=1, P=\frac{3}{2}, Q=0, R=\frac{1}{2}$.Then $2AP-BQ+R = 2(1)(\frac{3}{2}) - (1)(0) + \frac{1}{2} = 3 + \frac{1}{2} = \frac{7}{2}$.Wait,re-evaluating the partial fraction decomposition: $\frac{x^4}{(x-1)^2(x+1)} = x+1 + \frac{2x^2-x-1}{(x-1)^2(x+1)} = x+1 + \frac{(2x+1)(x-1)}{(x-1)^2(x+1)} = x+1 + \frac{2x+1}{(x-1)(x+1)}$.Using $\frac{2x+1}{(x-1)(x+1)} = \frac{3/2}{x-1} + \frac{1/2}{x+1}$.So $A=1, B=1, P=3/2, Q=0, R=1/2$. $2(1)(3/2) - 1(0) + 1/2 = 3.5 = 7/2$.

If $\frac{x^4}{(x-1)^2(x+1)}=Ax+B+\frac{P}{(x-1)}+\frac{Q}{(x-1)^2}+\frac{R}{x+1}$,then $2AP-BQ+R=$

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