If frac{x^4}{(x-1)(x-2)}=f(x)+frac{A}{x-1}+fr | vedclass

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(D) Given $\frac{x^4}{(x-1)(x-2)} = f(x) + \frac{A}{x-1} + \frac{B}{x-2}$.First,perform polynomial division for $\frac{x^4}{x^2-3x+2}$.$x^4 = (x^2-3x+

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$20$

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(D) Given $\frac{x^4}{(x-1)(x-2)} = f(x) + \frac{A}{x-1} + \frac{B}{x-2}$.First,perform polynomial division for $\frac{x^4}{x^2-3x+2}$.$x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$.So,$\frac{x^4}{(x-1)(x-2)} = x^2+3x+7 + \frac{15x-14}{(x-1)(x-2)}$.Using partial fractions for $\frac{15x-14}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.$15x-14 = A(x-2) + B(x-1)$.For $x=1$,$15-14 = A(1-2) \implies A = -1$.For $x=2$,$30-14 = B(2-1) \implies B = 16$.Thus,$f(x) = x^2+3x+7$.$f(-2) = (-2)^2 + 3(-2) + 7 = 4 - 6 + 7 = 5$.Therefore,$f(-2)+A+B = 5 + (-1) + 16 = 20$.

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

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