Question

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

Answer 1

`(2x)/(4x^2+12x+9)=(-3)/(2x+3)^2+1/(2x+3)`

Explanation:

Start with the given denominator `(4x^2+12x+9)`
this is equal to `(2x+3)^2`

so that the fraction is

`(2x)/(4x^2+12x+9)=(2x)/(2x+3)^2=A/(2x+3)^2+B/(2x+3)`

the LCD`=(2x+3)^2`

the equation becomes

`(2x)/(4x^2+12x+9)=(2x)/(2x+3)^2=A/(2x+3)^2+(B(2x+3))/(2x+3)^2`

and also

`(2x)/(2x+3)^2=(A+B(2x+3))/(2x+3)^2`

`(2x)/(2x+3)^2=(A+2Bx+3B)/(2x+3)^2`

rearranging

`(2x+0*x^0)/(2x+3)^2=(2Bx+(A+3B)*x^0)/(2x+3)^2`

the equations for the variables A, B are

`2B=2`
`A+3B=0`

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

final answer is

`(2x)/(4x^2+12x+9)=(2x)/(2x+3)^2=A/(2x+3)^2+B/(2x+3)`

`(2x)/(4x^2+12x+9)=(2x)/(2x+3)^2=(-3)/(2x+3)^2+1/(2x+3)`