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

Feb 7, 2016

$\frac{7}{x + 3} + \frac{4}{x - 2}$

#### Explanation:

First step is to factorise the denominator

${x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right)$

since these factors are linear then the numerators will be constants.

$\frac{11 x - 2}{\left(x + 3\right) \left(x - 2\right)} = \frac{A}{x + 3} + \frac{B}{x - 2}$

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 $\Rightarrow B = 4$

let x = -3 in (1): - 35 = - 5A $\Rightarrow A = 7$

$\Rightarrow \frac{11 x - 2}{{x}^{2} + x - 6} = \frac{7}{x + 3} + \frac{4}{x - 2}$