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

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

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$x-y+20=0$

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(B) Given,$\frac{x^2+5x+7}{(x-3)^3} = \frac{A}{x-3} + \frac{B}{(x-3)^2} + \frac{C}{(x-3)^3}$.Multiplying both sides by $(x-3)^3$,we get:$x^2+5x+7 = A(x-3)^2 + B(x-3) + C$  --- $(i)$Putting $x=3$ in $(i)$:$3^2 + 5(3) + 7 = C$ $\Rightarrow 9 + 15 + 7 = C$ $\Rightarrow C = 31$.Putting $x=0$ in $(i)$:$7 = A(-3)^2 + B(-3) + 31$ $\Rightarrow 9A - 3B = -24$ $\Rightarrow 3A - B = -8$ --- (ii)Putting $x=1$ in $(i)$:$1^2 + 5(1) + 7 = A(1-3)^2 + B(1-3) + 31$ $\Rightarrow 13 = 4A - 2B + 31$ $\Rightarrow 4A - 2B = -18$ $\Rightarrow 2A - B = -9$ --- (iii)Subtracting (iii) from (ii):$(3A - B) - (2A - B) = -8 - (-9) \Rightarrow A = 1$.Substituting $A=1$ in (iii):$2(1) - B = -9 \Rightarrow B = 11$.Thus,$A=1, B=11, C=31$.The equation of the line with slope $m=A=1$ passing through $(B, C) = (11, 31)$ is:$y - y_1 = m(x - x_1) \Rightarrow y - 31 = 1(x - 11)$$y - 31 = x - 11 \Rightarrow x - y + 20 = 0$.

If $\frac{x^2+5x+7}{(x-3)^3} = \frac{A}{x-3} + \frac{B}{(x-3)^2} + \frac{C}{(x-3)^3}$,then the equation of the line having slope $A$ and passing through the point $(B, C)$ is

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