If frac{x^2-3x+2}{(x-4)(x-3)^2}=frac{A}{x-4}+ | vedclass

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(C) Given the partial fraction decomposition: $\frac{x^2-3x+2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$.Multiply both sides

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$-1$

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(C) Given the partial fraction decomposition: $\frac{x^2-3x+2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$.Multiply both sides by $(x-4)(x-3)^2$ to get: $x^2-3x+2 = A(x-3)^2 + B(x-3)(x-4) + C(x-4)$.To find $A$,set $x=4$: $4^2 - 3(4) + 2 = A(4-3)^2 \Rightarrow 16 - 12 + 2 = A(1)^2 \Rightarrow A = 6$.To find $C$,set $x=3$: $3^2 - 3(3) + 2 = C(3-4) \Rightarrow 9 - 9 + 2 = C(-1) \Rightarrow 2 = -C \Rightarrow C = -2$.To find $B$,compare the coefficients of $x^2$ on both sides: $1 = A + B$. Since $A=6$,we have $1 = 6 + B \Rightarrow B = -5$.Finally,calculate $A+B+C = 6 + (-5) + (-2) = 6 - 7 = -1$.

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

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