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

Oct 17, 2016

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

#### Explanation:

To write the given expression into partial fractions we think about factorizing the denominator.

Let us factorize the denominator

$\textcolor{b l u e}{{x}^{3} - 2 {x}^{2} - x + 2}$
$= \textcolor{b l u e}{{x}^{2} \left(x - 2\right) - \left(x - 2\right)}$
$= \textcolor{b l u e}{\left(x - 2\right) \left({x}^{2} - 1\right)}$
Applying the identity of polynomials:
$\textcolor{\mathmr{and} a n \ge}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$
we have:
$\textcolor{b l u e}{{x}^{3} - 2 {x}^{2} - x + 2}$
$= \textcolor{b l u e}{\left(x - 2\right) \left({x}^{2} - {1}^{2}\right)}$
$= \textcolor{b l u e}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)}$

Let us decompose the rational expression by finding $A , B , \mathmr{and} C$

$\textcolor{b r o w n}{\frac{A}{x - 2} + \frac{B}{x - 1} + \frac{C}{x + 1}} = \textcolor{g r e e n}{\frac{3 x}{{x}^{3} - 2 {x}^{2} - x + 2}}$

$\textcolor{b r o w n}{\frac{A}{x - 2} + \frac{B}{x - 1} + \frac{C}{x + 1}}$

$= \textcolor{b r o w n}{\frac{A \left(x - 1\right) \left(x + 1\right)}{x - 2} + \frac{B \left(x - 2\right) \left(x + 1\right)}{x - 1} + \frac{C \left(x - 2\right) \left(x - 1\right)}{x + 1}}$

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

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

=(Ax^2-A+Bx^2-Bx-2B+Cx^2-3Cx+2C)/((x-2)(x-1)(x+1)

=color(brown)(((A+B+C)x^2+(-B-3C)x+(-A-2B+2C))/((x-2)(x-1)(x+1))

=color(brown)(((A+B+C)x^2+(-B-3C)x+(-A-2B+2C))/((x-2)(x-1)(x+1))=color(green)((3x)/(x^3-2x^2-x+2))
Then ,

$\Rightarrow \textcolor{b r o w n}{\left(A + B + C\right) {x}^{2} + \left(- B - 3 C\right) x + \left(- A - 2 B + 2 C\right)} = \textcolor{g r e e n}{3 x}$

We have a system of three equations with three unknowns $A , B \mathmr{and} C$

$A + B + C = 0$ eq1

$- B - 3 C = 3$ eq2

$- A - 2 B + 2 C = 0$ eq3

Starting to solve the system
eq2:$- B - 3 C = 3 \Rightarrow - B = 3 + 3 C \Rightarrow \textcolor{red}{B = - 3 - 3 C}$

Substituting $B$ in eq1 we have:

$A + B + C = 0$
$A - 3 - 3 C + C = 0 \Rightarrow A - 3 - 2 C = 0 \Rightarrow \textcolor{red}{A = 3 + 2 C}$

Substituting $B \mathmr{and} C$in eq3 we have:

$- A - 2 B + 2 C = 0$ eq3
$\Rightarrow - \left(\textcolor{red}{3 + 2 C}\right) - 2 \left(\textcolor{red}{- 3 - 3 C}\right) + 2 C = 0$
$\Rightarrow - 3 - 2 C + 6 + 6 C + 2 C = 0$
$\Rightarrow + 3 + 6 C = 0$
$\Rightarrow 6 C = - 3$
$\Rightarrow \textcolor{red}{C = - \frac{1}{2}}$

$\textcolor{red}{B = - 3 - 3 C} = - 3 - 3 \textcolor{red}{- \frac{1}{2}} = - 3 + \frac{3}{2}$
color(red)(B=-3/2
$\textcolor{red}{A = 3 + 2 C} = 3 + 2 \left(- \frac{1}{2}\right) = 3 - 1$
$\textcolor{red}{A = 2}$

Let us substitute the values:
$\textcolor{g r e e n}{\frac{3 x}{{x}^{3} - 2 {x}^{2} - x + 2}} = \textcolor{b r o w n}{\frac{\textcolor{red}{2}}{x - 2} + \frac{\textcolor{red}{- \frac{3}{2}}}{x - 1} + \frac{\textcolor{red}{\left(- \frac{1}{2}\right)}}{x + 1}}$

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