Question

How do you write the partial fraction decomposition of the rational expression `1/(4x^2 - 9)`?

Answer 1

`1/(4 x^2-9)= 1/(6(2 x-3)) -1/(6(2 x+3))`

Explanation:

`1/(4x^2-9) = 1/((2 x+3)(2 x-3))`

Let `1/((2 x+3)(2 x-3))= A/(2 x+3)+B/(2 x-3)` or

`1/cancel((2 x+3)(2 x-3))= (A(2 x-3)+B(2 x+3))/cancel((2 x+3)(2 x-3))` or

`A(2 x-3)+B(2 x+3) =1` Putting `x= 3/2` we get ,

`A(0)+ B(6) =1 :. 6 B = 1 :. B= 1/6` Putting `x= -3/2` we get ,

`A(-6)+ B(0) =1 :. -6 A = 1 :. A = -1/6`

`:.1/((2 x+3)(2 x-3))= (-1/6)/(2 x+3)+ (1/6)/(2 x-3)` or

`1/(4 x^2-9)= 1/(6(2 x-3)) -1/(6(2 x+3))` [Ans]