If $\frac{2 x^2+5 x+6}{(x+2)^3}=\frac{a}{x+2}+\frac{b}{(x+2)^2}+\frac{c}{(x+2)^3}$ then $\mathrm{a} . \mathrm{b}+\mathrm{b} . \mathrm{c}+\mathrm{c} . \mathrm{a}=$

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Given } \frac{2 x^2+5 x+6}{(x+2)^3}=\frac{a}{(x+2)}+\frac{b}{(x+2)^2}+\frac{c}{(x+3)^3} \ & \Rightarrow 2 x^2+5 x+6=a(x+2)^2+b(x+2)+c \ & \Rightarrow 2 x^2+5 x+6=x^2(a)+x(4 a+b)+(4 a+2 b+c) \end{aligned} $ Comparing we get, $ a=2, b=-3, c=4 $ Hence $ a \cdot b+b \cdot c+c \cdot a=-10 $

If $\frac{2 x^2+5 x+6}{(x+2)^3}=\frac{a}{x+2}+\frac{b}{(x+2)^2}+\frac{c}{(x+2)^3}$ then $\mathrm{a} .…

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