# How do you write the partial fraction decomposition of the rational expression  (2x+1)/(4x^2+12x-7)?

Dec 14, 2016

The answer is $= \frac{\frac{1}{4}}{2 x - 1} + \frac{\frac{3}{4}}{2 x + 7}$

#### Explanation:

Let's factorise the denominator

$4 {x}^{2} + 12 x - 7 = \left(2 x - 1\right) \left(2 x + 7\right)$

So, the decomposition into partial fractions is

$\frac{2 x + 1}{4 {x}^{2} + 12 x - 7} = \frac{2 x + 1}{\left(2 x - 1\right) \left(2 x + 7\right)}$

$= \frac{A}{2 x - 1} + \frac{B}{2 x + 7}$

$= \frac{A \left(2 x + 7\right) + B \left(2 x - 1\right)}{4 {x}^{2} + 12 x - 7}$

Therefore,

$2 x + 1 = A \left(2 x + 7\right) + B \left(2 x - 1\right)$

Let $x = \frac{1}{2}$, $\implies$, $2 = 8 A$, $\implies$, $A = \frac{1}{4}$

Let $x = - \frac{7}{2}$, $\implies$, $- 6 = - 8 B$, $\implies$, $B = \frac{3}{4}$

Therefore,

$\frac{2 x + 1}{4 {x}^{2} + 12 x - 7} = \frac{\frac{1}{4}}{2 x - 1} + \frac{\frac{3}{4}}{2 x + 7}$