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

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Answer 1 · 2016-02-18T08:50:01

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

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

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

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

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