# How do you express (2x) / (4x^2 + 12x + 9) in partial fractions?

Feb 18, 2016

$\frac{2 x}{4 {x}^{2} + 12 x + 9} = \frac{- 3}{2 x + 3} ^ 2 + \frac{1}{2 x + 3}$

#### Explanation:

Start with the given denominator $\left(4 {x}^{2} + 12 x + 9\right)$
this is equal to ${\left(2 x + 3\right)}^{2}$

so that the fraction is

$\frac{2 x}{4 {x}^{2} + 12 x + 9} = \frac{2 x}{2 x + 3} ^ 2 = \frac{A}{2 x + 3} ^ 2 + \frac{B}{2 x + 3}$

the LCD$= {\left(2 x + 3\right)}^{2}$

the equation becomes

$\frac{2 x}{4 {x}^{2} + 12 x + 9} = \frac{2 x}{2 x + 3} ^ 2 = \frac{A}{2 x + 3} ^ 2 + \frac{B \left(2 x + 3\right)}{2 x + 3} ^ 2$

and also

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

$\frac{2 x}{2 x + 3} ^ 2 = \frac{A + 2 B x + 3 B}{2 x + 3} ^ 2$

rearranging

$\frac{2 x + 0 \cdot {x}^{0}}{2 x + 3} ^ 2 = \frac{2 B x + \left(A + 3 B\right) \cdot {x}^{0}}{2 x + 3} ^ 2$

the equations for the variables A, B are

$2 B = 2$
$A + 3 B = 0$

Using Algebra to find
$B = 1$ and $A = - 3$

$\frac{2 x}{4 {x}^{2} + 12 x + 9} = \frac{2 x}{2 x + 3} ^ 2 = \frac{A}{2 x + 3} ^ 2 + \frac{B}{2 x + 3}$
$\frac{2 x}{4 {x}^{2} + 12 x + 9} = \frac{2 x}{2 x + 3} ^ 2 = \frac{- 3}{2 x + 3} ^ 2 + \frac{1}{2 x + 3}$