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

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

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Answer 1 · 2016-02-18T08:50:01

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

Explanation:

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

so that the fraction is

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

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

the equation becomes

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

and also

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

$\frac{2 x}{{(2 x + 3)}^{2}} = \frac{A + 2 B x + 3 B}{{(2 x + 3)}^{2}}$ $(2x)/(2x+3)^2=(A+2Bx+3B)/(2x+3)^2$

rearranging

$\frac{2 x + 0 \cdot x^{0}}{{(2 x + 3)}^{2}} = \frac{2 B x + (A + 3 B) \cdot x^{0}}{{(2 x + 3)}^{2}}$ $(2x+0*x^0)/(2x+3)^2=(2Bx+(A+3B)*x^0)/(2x+3)^2$

the equations for the variables A, B are

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

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

final answer is

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

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

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

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Explanation:

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