Question

How do you express `(6x^2+7x-6)/((x^2-4)(x+2))` in partial fractions?

Answer 1

`2/(x-2) + 4/(x+2) - 1/(x+2)^2`

Explanation:

first step here is to factor the denominator

`(x^2-4)(x+2) = (x-2)(x+2)(x+2) = (x-2)(x+2)^2`
Note also that the factors of `(x+2)^2 = (x+2)color(black)(" and")(x+2)^2`
Since these factors are linear , the numerators will be constants , say A , B and C.

`(6x^2+7x-6)/((x-2)(x+2)^2) = A/(x-2) + B/(x+2) + C/(x+2)^2`

multiply through by `(x-2)(x+2)^2`

the following equation will be referred to as ( 1)
`6x^2+7x-6 = A(x+2)^2 + B(x-2)(x+2) + C(x-2)`

The aim now is to find the values of A , B and C. Note that if x = 2 , the terms with B andC will be zero. If x = -2 , the terms with A and B will be zero. This is the starting point for finding values.

let x = 2 in (1) : 32 = 16A `rArr A = 4`

let x = -2 in (1) : 4 = - 4C `rArr C = - 1`

Any value for x , may be chosen to find B.

let x = 0 in (1) : - 6 = 4A - 4B - 2C

hence 4B = 4A - 2C + 6 = 8 + 2 + 6 = 16 `rArr B = 4`

`rArr (6x^2+7x-6)/((x^2-4)(x+2)) = 2/(x-2) + 4/(x+2) - 1/(x+2)^2`