# How do you write the partial fraction decomposition of the rational expression  (x+1)/(x^2(x-2))?

Feb 1, 2016

Explanation is given below

#### Explanation:

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

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

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

Equating the numerators

$x + 1 = A x \left(x - 2\right) + B \left(x - 2\right) + C {x}^{2}$

To make this easy for us, let us start by taking $x = 0$ and plug in the value, we get.

$0 + 1 = A \left(0\right) \left(0 - 2\right) + B \left(0 - 2\right) + C {\left(0\right)}^{2}$
$1 = 0 - 2 B + 0$

$- 2 B = 1$
$B = - \frac{1}{2}$

Note $x = 0$ helped us eliminate $A$ and $C$ now let us take $x = 2$
$2 + 1 = A \left(2\right) \left(2 - 2\right) + B \left(2 - 2\right) + C \left({2}^{2}\right)$
$3 = A \left(2\right) \left(0\right) + B \left(0\right) + 4 C$
$3 = 0 + 0 + 4 C$

$4 C = 3$
$C = \frac{3}{4}$

We have to find $A$ Let us plug in x=1
$1 + 1 = A \left(1\right) \left(1 - 2\right) + B \left(1 - 2\right) + C \left({1}^{2}\right)$
$2 = A \left(1\right) \left(- 1\right) - B + C$

$- A - B + C = 2$
$- A - \left(- \frac{1}{2}\right) + \frac{3}{4} = 2$

$- A + \frac{1}{2} + \frac{3}{4} = 2$

$- A + \frac{2}{4} + \frac{3}{4} = 2$
$- A + \frac{5}{4} = 2$

$- A = 2 - \frac{5}{4}$

$- A = \frac{8 - 5}{4}$

$- A = \frac{3}{4}$

$A = - \frac{3}{4}$

$\frac{x + 1}{{x}^{2} \left(x - 2\right)} = - \frac{3}{4 x} - \frac{1}{2 {x}^{2}} + \frac{3}{4 \left(x - 2\right)}$