# How do you write the partial fraction decomposition of the rational expression (2x^3+2x^2-9x-20)/((x^2-x-6)(x+2))?

Aug 1, 2017

$\frac{2 {x}^{3} + 2 {x}^{2} - 9 x - 20}{\left({x}^{2} - x - 6\right) \left(x + 2\right)} = 2 + \frac{1}{x - 3} - \frac{3}{x + 2} + \frac{2}{x + 2} ^ 2$

#### Explanation:

Before we express in partial fractions, ensure that the degree of numerator is less than that of denominator.

$\frac{2 {x}^{3} + 2 {x}^{2} - 9 x - 20}{\left({x}^{2} - x - 6\right) \left(x + 2\right)}$

= $\frac{2 {x}^{3} + 2 {x}^{2} - 9 x - 20}{{x}^{3} + {x}^{2} - 8 x - 12}$

= $2 + \frac{7 x + 4}{{x}^{3} + {x}^{2} - 8 x - 12}$

= $2 + \frac{7 x + 4}{\left(x - 3\right) \left(x + 2\right) \left(x + 2\right)}$

= $2 + \frac{7 x + 4}{\left(x - 3\right) {\left(x + 2\right)}^{2}}$

Now let $\frac{7 x + 4}{\left(x - 3\right) {\left(x + 2\right)}^{2}} \equiv \frac{A}{x - 3} + \frac{B}{x + 2} + \frac{C}{x + 2} ^ 2$

or $\frac{7 x + 4}{\left(x - 3\right) {\left(x + 2\right)}^{2}} = \frac{A {\left(x + 2\right)}^{2} + B \left(x - 3\right) \left(x + 2\right) + C \left(x - 3\right)}{\left(x - 3\right) {\left(x + 2\right)}^{2}}$

if $x = 3$, then $25 A = 25$ i.e. $A = 1$

if $x = - 2$, then $- 5 C = - 10$ i.e. $C = 2$

and comparing coefficients of ${x}^{2}$ on both sides

$A + B + C = 0$ i.e. $B = - 3$

Hence

$\frac{2 {x}^{3} + 2 {x}^{2} - 9 x - 20}{\left({x}^{2} - x - 6\right) \left(x + 2\right)} = 2 + \frac{1}{x - 3} - \frac{3}{x + 2} + \frac{2}{x + 2} ^ 2$