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

##### 1 Answer

#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#