Question

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

Answer 1

`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`= 1/3`then the term with A will be zero and if x `= -2/3` the term with B will be zero.

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 )`