Question

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

Answer 1

`1/x + 2/(x-1)`

Explanation:

First step is to factor the denominator

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

`rArr (3x-1)/(x(x-1)) = A/x + 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 `rArr A = 1`

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

`rArr (3x -1)/(x^2 - x ) = 1/x + 2/(x - 1 )`