# How do you write the partial fraction decomposition of the rational expression (3x^2 + 10x -5)/ ( (x+1)^2(x-2) )?

##### 1 Answer
Dec 30, 2016

The answer is $= \frac{4}{x + 1} ^ 2 + \frac{3}{x - 2}$

#### Explanation:

Let's do the decomposition into partial fractions

$\frac{3 {x}^{2} + 10 x - 5}{{\left(x + 1\right)}^{2} \left(x - 2\right)} = \frac{A}{x + 1} ^ 2 + \frac{B}{x + 1} + \frac{C}{x - 2}$

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

Therefore,

$3 {x}^{2} + 10 x - 5 = A \left(x - 2\right) + B \left(x + 1\right) \left(x - 2\right) + C {\left(x + 1\right)}^{2}$

Let $x = - 1$, $\implies$, $- 12 = - 3 A$, $\implies$, $A = 4$

Ler $x = 2$, $\implies$, $27 = 9 C$, $\implies$, $C = 3$

Coefficients of ${x}^{2}$

$3 = B + C$, $\implies$, $B = 3 - C = 0$

So,

$\frac{3 {x}^{2} + 10 x - 5}{{\left(x + 1\right)}^{2} \left(x - 2\right)} = \frac{4}{x + 1} ^ 2 + \frac{0}{x + 1} + \frac{3}{x - 2}$