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

Feb 7, 2016

$\frac{2}{3 x + 2} + \frac{1}{3 x - 1}$

#### Explanation:

first step here is to factor the denominator

$9 {x}^{2} + 3 x - 2 = \left(3 x + 2\right) \left(3 x - 1\right)$

since these factors are linear , the numerators will be constants

$\frac{9 x}{\left(3 x + 2\right) \left(3 x - 1\right)} = \frac{A}{3 x + 2} + \frac{B}{3 x - 1}$

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$= \frac{1}{3}$then the term with A will be zero and if x $= - \frac{2}{3}$ the term with B will be zero.

let $x = \frac{1}{3} \textcolor{b l a c k}{\text{ in}} \left(1\right) : 3 = 3 B \Rightarrow B = 1$

let $x = - \frac{2}{3} \textcolor{b l a c k}{\text{ in}} \left(1\right) : - 6 = - 3 A \Rightarrow A = 2$

$\Rightarrow \frac{9 x}{9 {x}^{2} + 3 x - 2} = \frac{2}{3 x + 2} + \frac{1}{3 x - 1}$