# How do you express (4x^3-22x^2+33x-6)/((x-3)^3(x+2)) in partial fractions?

Nov 8, 2016

The decomposition is $= \frac{\frac{3}{5}}{x - 3} ^ 3 + \frac{\frac{42}{25}}{x - 3} ^ 2 + \frac{\frac{308}{125}}{x - 3} + \frac{\frac{192}{125}}{x + 2}$

#### Explanation:

The decomposition is $\frac{4 {x}^{3} - 22 {x}^{2} + 33 x - 6}{{\left(x - 3\right)}^{3} \left(x + 2\right)} = \frac{A}{x - 3} ^ 3 + \frac{B}{x - 3} ^ 2 + \frac{C}{x - 3} + \frac{D}{x + 2}$
$= A \left(x + 2\right) + B \left(x - 3\right) \left(x + 2\right) + C {\left(x - 3\right)}^{2} \left(x + 2\right) + D {\left(x - 3\right)}^{3}$

$4 {x}^{3} - 22 {x}^{2} + 33 x - 6 = A \left(x + 2\right) + B \left(x - 3\right) \left(x + 2\right) + C {\left(x - 3\right)}^{2} \left(x + 2\right) + D {\left(x - 3\right)}^{3}$
let $x = 3$$\implies$$5 A = 3$$\implies$$A = \frac{3}{5}$
let $x = - 2$$\implies$$- 125 D = - 192$$\implies$$D = \frac{192}{125}$
coefficients of ${x}^{3}$$\implies$$C + D = 4$$\implies$$C = 4 - \frac{192}{125} = \frac{308}{125}$
and putting $x = 0$, we get $- 6 = 2 A - 6 B + 18 C - 27 D$

or $\implies 6 B = 6 + \frac{6}{5} + 18 \cdot \frac{308}{25} - 27 \cdot \frac{192}{125}$

= $\frac{36}{5} + 9 \left(\frac{616}{125} - \frac{576}{25}\right)$

= $\frac{36}{5} + \frac{360}{125}$

= $\frac{180}{25} + \frac{72}{25} = \frac{252}{25}$

$\implies B = \frac{42}{25}$