# How do you express  (5x^2+3x-2)/(x^3+2x^2)  in partial fractions?

Oct 27, 2016

The result is
$\frac{5 {x}^{2} + 3 x - 2}{{x}^{3} + 2 {x}^{2}} = - \frac{1}{x} ^ 2 + \frac{2}{x} + \frac{3}{x + 2}$

#### Explanation:

We need to simplify the denominator
${x}^{3} + 2 {x}^{2} = {x}^{2} \left(x + 2\right)$

Let $\frac{5 {x}^{2} + 3 x - 2}{{x}^{3} + 2 {x}^{2}} = \frac{5 {x}^{2} + 3 x - 2}{{x}^{2} \left(x + 2\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x + 2}$
$= \frac{A \left(x + 2\right) + B x + C {x}^{2}}{{x}^{2} \left(x + 2\right)}$
So $5 {x}^{2} + 3 x - 2 = A \left(x + 2\right) + B x \left(x + 2\right) + C {x}^{2}$

Let $x = 0$ then $- 2 = 2 A$ $\implies$$A = - 1$

Coefficients of ${x}^{2}$
$5 = B + C$

Coefficients of $x$
$3 = A + 2 B$ $\implies$ $3 = - 1 + 2 B$ $\implies$$B = 2$

and $5 = 2 + C$ $\implies$$C = 3$

so the result is
$\frac{5 {x}^{2} + 3 x - 2}{{x}^{3} + 2 {x}^{2}} = - \frac{1}{x} ^ 2 + \frac{2}{x} + \frac{3}{x + 2}$