# How do you write the partial fraction decomposition of the rational expression (5x^3 -9x+6)/(x^3-x^2)?

Apr 26, 2017

$\frac{5 {x}^{3} - 9 x + 6}{{x}^{3} - {x}^{2}} = 5 - \frac{6}{x} + \frac{1}{x - 1} + \frac{10}{x + 1}$

#### Explanation:

$\frac{5 {x}^{3} - 9 x + 6}{{x}^{3} - {x}^{2}}$

= $\frac{5 \left({x}^{3} - {x}^{2}\right) + 5 {x}^{2} - 9 x + 6}{x \left(x - 1\right) \left(x + 1\right)}$

= $5 + \frac{5 {x}^{2} - 9 x + 6}{x \left(x - 1\right) \left(x + 1\right)}$

What we have done above is taken something out to make the degree of numerator less than that of denominator and factorizedenominator. Let us now convert this part into partial fractions.

Let $\frac{5 {x}^{2} - 9 x + 6}{x \left(x - 1\right) \left(x + 1\right)} \equiv \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$

or $\frac{5 {x}^{2} - 9 x + 6}{x \left(x - 1\right) \left(x + 1\right)} = A \left(x - 1\right) \left(x + 1\right) + B x \left(x + 1\right) + C x \left(x - 1\right)$

Putting $x = 0$ in this we get $- A = 6$ or $A = - 6$

putting $x = - 1$, we get $2 C = 20$ or $C = 10$ and

putting $x = 1$, we get $2 B = 2$ or $B = 1$

Hence, $\frac{5 {x}^{3} - 9 x + 6}{{x}^{3} - {x}^{2}}$

= $5 - \frac{6}{x} + \frac{1}{x - 1} + \frac{10}{x + 1}$