# How do you express 1/ [ (x-1)(x+2)(x-3)]  in partial fractions?

Jun 25, 2018

Shown below

#### Explanation:

$\frac{1}{\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)}$

$= \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 3}$

$\implies 1 = A \left(x + 2\right) \left(x - 3\right) + B \left(x - 1\right) \left(x - 3\right) + C \left(x - 1\right) \left(x + 2\right)$

Let $r = - 2$ Solving to give $B = \frac{1}{15}$

Let $r = 3$ Solving to give $C = \frac{1}{10}$

Let $r = 1$ Solving to give $A = - \frac{1}{6}$

Hence our new partial fraction form is...

$= - \frac{1}{6 \left(x - 1\right)} + \frac{1}{15 \left(x + 2\right)} + \frac{1}{10 \left(x - 3\right)}$