Resolve `(16)/((x-2)(x+2)^(2))` into partial fractions.

1.	Resolve `(16)/((x-2)(x+2)^(2))` into partial fractions.
Answer» `Let (16)/((x-2)(x+2)^(2))=(A)/(x-2)+(B)/(x+2)+( C)/((x+2)^(2))` `or (16)/((x-2)(x+2)^(2))=(A(x+2)^(2)+B(x-2)(x+2)+C(x-2))/((x-2)(x+2)^(2))` `therefore 16-= A(x+2)^(2)+b(x-2)(x+2)+C(x-2) . . . .(i)` `or 16-=(A+B)x^(2)+(4A+C)x+(4A-4B-2C).. . . .(ii)` Putting `(x-2)=0 or x=2` in (i), we get `A=1.` Putting `(x+2)=0 or x=-2` in (i), we get `C=-4.` Comparing the coefficients of `x^(2)` on both sides of (ii) , we get `A+B=0 or B=-A=-1` Thus `A=1,B=-1 and C=-4.` `therefore (16)/((x-2)(x+2)^(2))=[(1)/((x-2))-(1)/((x+2))-(4)/((x+2)^(2))].`

Discussion

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