# How do you express  (x^2 + 2x) / (x^3 - x^2 + x - 1)  in partial fractions?

##### 1 Answer
Dec 22, 2016

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

#### Explanation:

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

$\textcolor{w h i t e}{\frac{{x}^{2} + 2 x}{{x}^{3} - {x}^{2} + x - 1}} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 1}$

$\textcolor{w h i t e}{\frac{{x}^{2} + 2 x}{{x}^{3} - {x}^{2} + x - 1}} = \frac{A \left({x}^{2} + 1\right) + \left(B x + C\right) \left(x - 1\right)}{{x}^{3} - {x}^{2} + x - 1}$

$\textcolor{w h i t e}{\frac{{x}^{2} + 2 x}{{x}^{3} - {x}^{2} + x - 1}} = \frac{\left(A + B\right) {x}^{2} + \left(- B + C\right) x + \left(A - C\right)}{{x}^{3} - {x}^{2} + x - 1}$

Equating coefficients, we find:

$\left\{\begin{matrix}A + B = 1 \\ - B + C = 2 \\ A - C = 0\end{matrix}\right.$

Adding all three of these equations, the $B$ and $C$ terms cancel out, leaving us with:

$2 A = 3$

Hence $A = \frac{3}{2}$

Then from the first equation, we find:

$B = 1 - A = 1 - \frac{3}{2} = - \frac{1}{2}$

From the third equation, we find:

$C = A = \frac{3}{2}$

So:

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