Question

resolve 2x+3/xsquare-2x-3 into pattial fractions

Answer 1

(2x + 3)/(x² - 2x - 3) = (2x + 3)/(x² - 3x + x - 3)

                        = (2x + 3)/[x(x - 3) + 1(x - 3)]

                       = (2x + 3)/(x - 3)(x + 1)       ...(1)

So value of (2x + 3)/(x - 3)(x + 1) using partial fractions, we get

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

2x + 3 = A(x + 1) + B(x - 3)            ...(2)

Put x = -1 in (1) , we get

-2 + 3 = A(0) + B(-4)

⇒ 1 = -4B

⇒ B = -1/4

Again put x = 3 in (2), we get

6 + 3 = A(4) + B(0)

⇒ 9 = 4A

⇒ A = 9/4

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

(2x + 3)/(x² - 2x - 3) = (2x + 3)/(x - 3)(x + 1) = 9/4(x - 3)-1/4(x + 1)

Hence value of  (2x + 3)/(x² - 2x - 3) using partial fractions is

(2x + 3)/(x² - 2x - 3) = 9/4(x - 3)-1/4(x + 1)