# How do you express # (11x-2) /( x^2 + x-6)# in partial fractions?

##### 1 Answer

Feb 7, 2016

#7/(x+ 3 ) + 4/(x - 2 )#

#### Explanation:

First step is to factorise the denominator

# x^2 + x - 6 = (x+3)(x - 2 ) # since these factors are linear then the numerators will be constants.

# (11x - 2 )/((x + 3 )(x - 2 )) = A/(x + 3 ) + B/(x - 2 ) # multiply both sides by (x + 3 )(x - 2 )

hence : 11x - 2 = A(x - 2 ) + B(x + 3 )..............(1)

Task now is to find A and B .Note that if x = 2 , the term with A will be zero and if x = - 3 the term with B will be zero.

let x = 2 in (1) : 20 = 5B

# rArr B = 4# let x = -3 in (1): - 35 = - 5A

#rArr A = 7#

#rArr (11x - 2)/(x^2 + x - 6 ) = 7/(x + 3 ) + 4/(x - 2 ) #