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

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Answer 1 · 2016-12-14T08:47:26

The answer is $=(1/4)/(2x-1)+(3/4)/(2x+7)$

Let's factorise the denominator

$4x^2+12x-7=(2x-1)(2x+7)$

So, the decomposition into partial fractions is

$(2x+1)/(4x^2+12x-7)=(2x+1)/((2x-1)(2x+7))$

$=A/(2x-1)+B/(2x+7)$

$=(A(2x+7)+B(2x-1))/(4x^2+12x-7)$

Therefore,

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

Let $x=1/2$ , $=>$ , $2=8A$ , $=>$ , $A=1/4$

Let $x=-7/2$ , $=>$ , $-6=-8B$ , $=>$ , $B=3/4$

Therefore,

$(2x+1)/(4x^2+12x-7)=(1/4)/(2x-1)+(3/4)/(2x+7)$

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