# How do you express (x^3-x^2-5x) /( x^2-3x+2) in partial fractions?

Jul 19, 2018

The answer is $= x + 2 + \frac{5}{x - 1} - \frac{6}{x - 2}$

#### Explanation:

As the degree of the numerator is greater than the degree of the denominator, perform a long division first

$\textcolor{w h i t e}{a a a a}$${x}^{3} - {x}^{2} - 5 x + 0$$\textcolor{w h i t e}{a a a a}$$|$${x}^{2} - 3 x + 2$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 3 {x}^{2} + 2 x$$\textcolor{w h i t e}{a a a a a a a}$$|$$x + 2$

$\textcolor{w h i t e}{a a a a}$$0 {x}^{3} + 2 {x}^{2} - 7 x + 0$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 2 {x}^{2} - 6 x + 4$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 0 {x}^{2} - x - 4$

Therefore,

$\frac{{x}^{3} - {x}^{2} - 5 x}{{x}^{2} - 3 x + 2} = x + 2 - \frac{x + 4}{{x}^{2} - 3 x + 2}$

Perform the decomposition into partial fractions

$\frac{x + 4}{{x}^{2} - 3 x + 2} = \frac{x + 4}{\left(x - 1\right) \left(x - 2\right)}$

$= \frac{A}{x - 1} + \frac{B}{x - 2}$

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

Compare the numerators

$x + 4 = A \left(x - 2\right) + B \left(x - 1\right)$

Let $x = 1$, $\implies$, $5 = - A$

Let $x = 2$, $\implies$, $6 = B$

Finally,

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

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

#### Explanation:

Given rational function:

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

$= x + 2 - \setminus \frac{x + 4}{{x}^{2} - 3 x + 2}$

$= x + 2 - \setminus \frac{x + 4}{\left(x - 1\right) \left(x - 2\right)}$

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

Comparing corresponding coefficients to find $A = - 5$ & $B = 6$

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

$= x + 2 + \frac{5}{x - 1} - \frac{6}{x - 2}$