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

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(B) Given $\frac{x^{4}}{(x - 1)(x - 2)} = f(x) + \frac{A}{x - 1} + \frac{B}{x - 2}$.By performing polynomial long division,we divide $x^{4}$ by $(x -

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$f(x) = x^{2} + 3x + 7$

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(B) Given $\frac{x^{4}}{(x - 1)(x - 2)} = f(x) + \frac{A}{x - 1} + \frac{B}{x - 2}$.By performing polynomial long division,we divide $x^{4}$ by $(x - 1)(x - 2) = x^{2} - 3x + 2$.$x^{4} = (x^{2} - 3x + 2)(x^{2} + 3x + 7) + (15x - 14)$.Thus,$\frac{x^{4}}{(x - 1)(x - 2)} = x^{2} + 3x + 7 + \frac{15x - 14}{(x - 1)(x - 2)}$.We express $\frac{15x - 14}{(x - 1)(x - 2)}$ as partial fractions: $\frac{15x - 14}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$.$15x - 14 = A(x - 2) + B(x - 1)$.For $x = 1$: $15(1) - 14 = A(1 - 2) \Rightarrow 1 = -A \Rightarrow A = -1$.For $x = 2$: $15(2) - 14 = B(2 - 1) \Rightarrow 16 = B$.Comparing this with the given equation,$f(x) = x^{2} + 3x + 7$,$A = -1$,and $B = 16$.Checking the options: $f(x) = x^{2} + 3x + 7$ is option $B$,and $A + B = -1 + 16 = 15$ (not $17$),$A - B = -1 - 16 = -17$ (not $-18$).Thus,the correct option is $B$.

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

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