If frac{2x^2-3x+5}{(x-7)^3}=frac{A}{x-7}+frac | vedclass

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(C) Given $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$.Multiplying both sides by $(x-7)^3$,we get:$2x^2-3x+5 = A(x-7)

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) Given $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$.Multiplying both sides by $(x-7)^3$,we get:$2x^2-3x+5 = A(x-7)^2 + B(x-7) + C$$2x^2-3x+5 = A(x^2-14x+49) + Bx - 7B + C$$2x^2-3x+5 = Ax^2 + (B-14A)x + (49A-7B+C)$Comparing the coefficients of $x^2$,$x$,and the constant term:$A = 2$$B-14A = -3$ $\Rightarrow B-14(2) = -3$ $\Rightarrow B-28 = -3$ $\Rightarrow B = 25$$49A-7B+C = 5$ $\Rightarrow 49(2)-7(25)+C = 5$ $\Rightarrow 98-175+C = 5$ $\Rightarrow C-77 = 5$ $\Rightarrow C = 82$Now,calculate $2A-3B+C$:$2(2)-3(25)+82 = 4-75+82 = 11$.Therefore,option $C$ is correct.

If $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$,then $2A-3B+C=$

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