# How do I decompose the rational expression (x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) into partial fractions?

Mar 3, 2018

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

$= {x}^{2} + \frac{3}{x - 2} + \frac{- x - 1}{{x}^{2} + 1} , \mathmr{and} ,$

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

#### Explanation:

Observe that, the degree of the poly. in Nr. is more

than that of the Dr., making the given rational fun.

improper.

To make it proper, the long division is usually performed.

But, here we have a short-cut method to proceed :

${x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5$,

$= {x}^{5} - 2 {x}^{4} + {x}^{3} - 2 {x}^{2} + \left(2 {x}^{2} + x + 5\right)$,

$= {x}^{2} \left({x}^{3} - 2 {x}^{2} + x - 2\right) + \left(2 {x}^{2} + x + 5\right)$.

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

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

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

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

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

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

Thus, it remains to decompose $\frac{2 {x}^{2} + x + 5}{\left(x - 2\right) \left({x}^{2} + 1\right)}$

into partial fraction. This has been carried out as under :

We suppose, for some $A , B , C \in \mathbb{R}$,

$\frac{2 {x}^{2} + x + 5}{\textcolor{red}{\left(x - 2\right)} \left({x}^{2} + 1\right)} = \textcolor{red}{\frac{A}{x - 2}} + \frac{B x + C}{{x}^{2} + 1} \ldots \ldots \ldots \ldots \left(\star\right)$.

To determine the consts. $A , B , C$, we use Heaviside's Method :

$\textcolor{red}{A} = {\left[\frac{2 {x}^{2} + x + 5}{{x}^{2} + 1}\right]}_{\textcolor{red}{\left(x = 2\right)}} = \frac{2 \cdot {2}^{2} + 2 + 5}{{2}^{2} + 1} = 3$.

Sub.ing $A = 3$ in $\left(\star\right)$, we get,

$\frac{B x + C}{{x}^{2} + 1} = \frac{2 {x}^{2} + x + 5}{\left(x - 2\right) \left({x}^{2} + 1\right)} - \frac{3}{x - 2}$,

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

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

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

$\Rightarrow \frac{B x + C}{{x}^{2} + 1} = \frac{- x - 1}{{x}^{2} + 1}$.

$\therefore B = - 1 , C = - 1$.

Altogether, we have,

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

$= {x}^{2} + \frac{3}{x - 2} + \frac{- x - 1}{{x}^{2} + 1} , \mathmr{and} ,$

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