# How do you express #(9x)/(9x^2+3x-2)# in partial fractions?

##### 1 Answer

Feb 7, 2016

# 2/(3x + 2 ) + 1/(3x - 1 ) #

#### Explanation:

first step here is to factor the denominator

# 9x^2 + 3x - 2 = (3x + 2 )(3x - 1 ) # since these factors are linear , the numerators will be constants

#( 9x)/((3x + 2 )(3x - 1 )) = A/(3x +2 ) + B/(3x - 1 ) # now multiply through by (3x + 2 )(3x - 1 )

hence : 9x = A(3x - 1 ) + B(3x + 2 ).......................(1)

the next step is to find values for A and B. Note that if x

let

# x = 1/3 color(black)(" in") (1) : 3 = 3B rArr B = 1 # let

# x = -2/3color(black)(" in") (1) : -6 = -3A rArr A = 2#

# rArr (9x)/(9x^2 + 3x - 2 ) = 2/(3x + 2 ) + 1/(3x - 1 )#