# How do you write the partial fraction decomposition of the rational expression x/ (8x^2−6x+1)?

Aug 19, 2016

$\frac{x}{8 {x}^{2} - 6 x + 1} = \frac{\frac{1}{2}}{2 x - 1} - \frac{\frac{1}{2}}{4 x - 1}$.

#### Explanation:

The $D r . = 8 {x}^{2} - 6 x + 1 = \left(2 x - 1\right) \left(4 x - 1\right)$

Both the factors are linear polys. Hence, the reqd. decomposition

is given by,

$\frac{x}{8 {x}^{2} - 6 x + 1} = \frac{A}{2 x - 1} + \frac{B}{4 x - 1} , w h e r e , A , B \in \mathbb{R}$.

To determine A,&,B, Heavyside's Cover-up Method is useful :-

$A = {\left[\frac{x}{4 x - 1}\right]}_{x = \frac{1}{2}} = \frac{1}{2} / \left(4 \cdot \frac{1}{2} - 1\right) = \frac{1}{2}$.

$B = {\left[\frac{x}{2 x - 1}\right]}_{x = \frac{1}{4}} = \frac{1}{4} / \left(2 \cdot \frac{1}{4} - 1\right) = \frac{1}{4} / \left(- \frac{1}{2}\right) = - \frac{1}{2}$.

Therefore,

$\frac{x}{8 {x}^{2} - 6 x + 1} = \frac{\frac{1}{2}}{2 x - 1} - \frac{\frac{1}{2}}{4 x - 1}$.

Method II :-

When there are only two linear factors in $D r .$, this decomposition can easily be derived as under :

Observe that, $\left(4 x - 1\right) - \left(2 x - 1\right) = 2 x$

$\Rightarrow N r . = x = \frac{1}{2} \left\{\left(4 x - 1\right) - \left(2 x - 1\right)\right\}$. Therefore,

$\frac{x}{8 {x}^{2} - 6 x + 1} = \frac{1}{2} \left[\frac{\left(4 x - 1\right) - \left(2 x - 1\right)}{\left(4 x - 1\right) \left(2 x - 1\right)}\right]$

$= \frac{1}{2} \left[\frac{\cancel{4 x - 1}}{\cancel{4 x - 1} \left(2 x - 1\right)} - \frac{\cancel{2 x - 1}}{\left(4 x - 1\right) \cancel{2 x - 1}}\right]$

$= \frac{1}{2} \left[\frac{1}{2 x - 1} - \frac{1}{4 x - 1}\right]$, as obtained earlier!

Enjoy Maths.!