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 ) #