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

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Aicharya · Accepted Answer

Given, $\begin{aligned} & \frac{8}{(x+3)^2(x-2)}=\frac{A x+B}{(x+3)^2}+\frac{C}{x-2} \ & \Rightarrow \frac{8}{(x+3)^2(x-2)}=\frac{(A x+B)(x-2)+C(x+3)^2}{(x+3)^2(x-4)} \ & \Rightarrow \quad 8=(A x+B)(x-2)+C(x+3)^2 \ & \text {At } x=2 \ & \Rightarrow \quad 8=0+C(5)^2 \ & \Rightarrow \quad 8=25 C \Rightarrow C=\frac{8}{25} \ & \text {At } x=0 \text {, } \ & \Rightarrow \quad 8=(B)(-2)+C(9) \ & \Rightarrow \quad 8=-2 B+9 C \ & \Rightarrow \quad 8=-2 B+\frac{72}{25} \quad\left(\because C=\frac{8}{25}\right) \ & \Rightarrow \quad 2 B=\frac{72}{25}-8 \ & \Rightarrow \quad 2 B=\frac{72-200}{25}=\frac{-128}{25} \ & \Rightarrow \quad B=\frac{-64}{25} \ & \text {At } \quad x=1 \text {, } \ & \Rightarrow \quad 8=(A+B)(-1)+C(4)^2 \ & \Rightarrow \quad 8=-A-B+16 C \ & \Rightarrow \quad 8=-A+\frac{64}{25}+\frac{128}{25} \ & \Rightarrow \quad A=\frac{192}{25}-8 \Rightarrow A=\frac{-8}{25} \end{aligned}$ Now, $25(B+8 C-A)=25\left(\frac{-64}{25}+\frac{64}{25}+\frac{8}{25}\right)$ $=25 \times \frac{8}{25}=8$

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

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