How do you express (3x-1)/(x^2-x) in partial fractions?

1 Answer
Feb 19, 2016

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

Explanation:

First step is to factor the denominator

hence : ${x}^{2} - x = x \left(x - 1\right)$
Since the factors are linear then the numerators of the partial fractions will be constants , say A and B.

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

now multiply through by x(x-1)

3x - 1 = A(x-1) + Bx ...........................................(1)

The aim now is to calculate values of A and B. Note that if x = 0 the term with B will be zero and if x = 1 the term with A will be zero. This is the starting point in calculating values.

let x = 0 in (1) : - 1 = - A $\Rightarrow A = 1$

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

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