Question

Resolve into partial fraction for the following:

`(2x^2 - 5x - 7)/(x - 2)^3`

Answer 1

Let `(2x^2 - 5x - 7)/(x - 2)^3 = "A"/(x - 2) + "B"/(x - 2)^2 + "C"/(x - 2)^3`  ....(1)

`(2x^2 - 5x - 7)/(x - 2)^3 = ("A"(x - 2)^2 + "B"(x - 2) + "C")/(x - 2)^3`

2x² – 5x – 7 = A(x – 2)² + B(x – 2) + C

2x² – 5x – 7 = A(x² – 4x + 4) + B(x – 2) + C …….. (2)

Put x = 2 in (2) we get

2(2²) – 5(2) – 7 = A(0) + B(0) + C

8 – 10 – 7 = 0 + 0 + C

-9 = C

C = -9

Equating coefficient of x2 on both sides of (2) we get

2 = A

A = 2

Equating coefficient of x on both sides of (2) we get

-5 = A(-4) + B(1)

-5 = 2(-4) + B(∵ A = 2)

-5 = -8 + B

B = 8 – 5 = 3

Using A = 2, B = 3, C = -9 in (1) we get

`(2x^2 - 5x - 7)/(x - 2)^3 = 2/(x - 2) + 3/(x - 2)^2 + (-9)/(x - 2)^3`

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