If frac{8}{(x+3)^2(x-2)}=frac{Ax+B}{(x+3)^2}+ | vedclass

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(C) Given,$\frac{8}{(x+3)^2(x-2)}=\frac{Ax+B}{(x+3)^2}+\frac{C}{x-2}$ Multiplying both sides by $(x+3)^2(x-2)$,we get: $8 = (Ax+B)(x-2) + C(x+3)^2$ Se

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$8$

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(C) Given,$\frac{8}{(x+3)^2(x-2)}=\frac{Ax+B}{(x+3)^2}+\frac{C}{x-2}$ Multiplying both sides by $(x+3)^2(x-2)$,we get: $8 = (Ax+B)(x-2) + C(x+3)^2$ Setting $x=2$: $8 = 0 + C(2+3)^2 \Rightarrow 8 = 25C \Rightarrow C = \frac{8}{25}$ Setting $x=-3$: $8 = (A(-3)+B)(-3-2) + 0 \Rightarrow 8 = (-3A+B)(-5) \Rightarrow 8 = 15A - 5B$ Comparing coefficients of $x^2$: $0 = A + C \Rightarrow A = -C = -\frac{8}{25}$ Substituting $A$ into $8 = 15A - 5B$: $8 = 15(-\frac{8}{25}) - 5B \Rightarrow 8 = -\frac{24}{5} - 5B \Rightarrow 5B = -\frac{24}{5} - 8 = -\frac{64}{5} \Rightarrow B = -\frac{64}{25}$ Now,calculate $25(B+8C-A)$: $25(-\frac{64}{25} + 8(\frac{8}{25}) - (-\frac{8}{25})) = 25(-\frac{64}{25} + \frac{64}{25} + \frac{8}{25}) = 25(\frac{8}{25}) = 8$

If $\frac{8}{(x+3)^2(x-2)}=\frac{Ax+B}{(x+3)^2}+\frac{C}{x-2}$,then $25(B+8C-A)=$

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