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

Mar 8, 2018

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

#### Explanation:

We need to write these in terms of each factors.

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

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

Putting in $x = - 2$:
${\left(- 2\right)}^{2} = A \left(- 2 + 2\right) + B \left(- 2 - 1\right)$
$4 = - 3 B$
$B = - \frac{4}{3}$

Putting in $x = 1$:
${1}^{2} = A \left(1 + 2\right) + B \left(1 - 1\right)$
$1 = 3 A$
$A = \frac{1}{3}$

${x}^{2} / \left(\left(x - 1\right) \left(x + 2\right)\right) = \frac{\frac{1}{3}}{x - 1} + \frac{- \frac{4}{3}}{x + 2}$
$\textcolor{w h i t e}{{x}^{2} / \left(\left(x - 1\right) \left(x + 2\right)\right)} = \frac{1}{3 \left(x - 1\right)} - \frac{4}{3 \left(x + 2\right)}$

Mar 8, 2018

$1 + \frac{1}{3} \cdot \frac{1}{x - 1} - \frac{4}{3} \cdot \frac{1}{x + 2}$

#### Explanation:

${x}^{2} / \left[\left(x - 1\right) \left(x + 2\right)\right]$

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

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

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

Now, I decomposed fraction into basic ones,

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

After expanding denominator,

$A \cdot \left(x + 2\right) + B \cdot \left(x - 1\right) = x - 2$

Set $x = - 2$, $- 3 B = - 4$, so $B = \frac{4}{3}$

Set $x = 1$, $3 A = - 1$, so $A = - \frac{1}{3}$

Hence,

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

Thus,

${x}^{2} / \left[\left(x - 1\right) \left(x + 2\right)\right]$

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

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