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

Mar 5, 2016

x/(x^3−x^2−2x+2)hArrx/(x^2-2)-1/(x-1)

#### Explanation:

To express x/(x^3−x^2−2x+2) in partial fractions, we should first factorize (x^3−x^2−2x+2).

(x^3−x^2−2x+2) = ${x}^{2} \left(x - 1\right) - 2 \left(x - 1\right) = \left({x}^{2} - 2\right) \left(x - 1\right)$

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

The latter can be expanded as $\frac{\left(A x + B\right) \left(x - 1\right) + C \left({x}^{2} - 2\right)}{\left({x}^{2} - 2\right) \left(x - 1\right)}$ or

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

$\frac{\left(A + C\right) {x}^{2} + \left(B - A\right) x - B - 2 C}{\left({x}^{2} - 2\right) \left(x - 1\right)}$ and as it is equal to x/(x^3−x^2−2x+2), we have

$A + C = 0$, $B - A = 1$ and $- B - 2 C = 0$

Eliminating C from first and third equation, by multiplying first by $2$ and adding it to third, we get $2 A - B = 0$.

Adding this to second, we get $A = 1$. This gives us $B = 0$ and then $C = - 1$.

Hence x/(x^3−x^2−2x+2)hArrx/(x^2-2)-1/(x-1)