# How do you write the partial fraction decomposition of the rational expression  (x+13)/(x^3+2x^2-5x-6)?

Mar 26, 2017

The answer is $= \frac{1}{x - 2} - \frac{2}{x + 1} + \frac{1}{x + 3}$

#### Explanation:

We need to factorise the denominator

Let $f \left(x\right) = {x}^{3} + 2 {x}^{2} - 5 x - 6$

Then

$f \left(2\right) = 8 + 8 - 10 - 6 = 0$

Therefore,

$\left(x - 2\right)$ is a factor of $f \left(x\right)$

We can do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} + 2 {x}^{2} - 5 x - 6$$\textcolor{w h i t e}{a a a a}$$|$$x - 2$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$|$${x}^{2} + 4 x + 3$

$\textcolor{w h i t e}{a a a a a}$$0 + 4 {x}^{2} - 5 x$

$\textcolor{w h i t e}{a a a a a a a}$$+ 4 {x}^{2} - 8 x$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 0 + 3 x - 6$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$+ 3 x - 6$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$+ 0 - 0$

so,

${x}^{3} + 2 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left({x}^{2} + 4 x + 3\right)$

$= \left(x - 2\right) \left(x + 1\right) \left(x + 3\right)$

Now, we perform the decomposition into partial fractions

$\frac{x + 13}{{x}^{3} + 2 {x}^{2} - 5 x - 6} = \frac{x + 13}{\left(x - 2\right) \left(x + 1\right) \left(x + 3\right)}$

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

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

The denominators are the same, we compare the numerators

$x + 13 = A \left(x + 1\right) \left(x + 3\right) + B \left(x - 2\right) \left(x + 3\right) + C \left(x - 2\right) \left(x + 1\right)$

Let, $x = 2$, $\implies$,$15 = 15 A$, $\implies$,$A = 1$

Let, $x = - 1$, $\implies$, $12 = - 6 B$, $\implies$, $B = - 2$

Let, $x = - 3$, $\implies$, $10 = 10 C$, $\implies$, $C = 1$

So,

$\frac{x + 13}{{x}^{3} + 2 {x}^{2} - 5 x - 6} = \frac{1}{x - 2} - \frac{2}{x + 1} + \frac{1}{x + 3}$