How do you write the partial fraction decomposition of the rational expression #(3x^2+2x)/(x^2-4)#?

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Answer 1 · 2016-11-26T16:47:27

The answer is #=3-2/(x+2)+4/(x-2)#

As the degree of the numerator is #=# the degree of the denominator

Let's do a long division

#color(white)(aaaaa)##3x^2+2x##color(white)(aaaaa)##∣##x^2-4#

#color(white)(aaaaa)##3x^2-12##color(white)(aaaaa)##∣##3#

#color(white)(aaaaaa)##0+2x+12#

#(3x^2+2x)/(x^2-4)=3+(2x+12)/(x^2-4)#

Let's factorise the denominator

x^2-4=(x+2)(x-2)#

Let's do the decomposition in partial fractions

#(2x+12)/(x^2-4)=(2x+12)/((x+2)(x-2))=A/(x+2)+B/(x-2)#

#=(A(x-2)+B(x+2))/((x+2)(x-2))#

So, #(2x+12)=(A(x-2)+B(x+2))#

Let #x=2#, #16=4B#, #=>#, #B=4#

Let #x=-2#, #8=-4A#, #=>#, #A=-2#

So, #(2x+12)/(x^2-4)=-2/(x+2)+4/(x-2)#

And finally we have

#(3x^2+2x)/(x^2-4)=3-2/(x+2)+4/(x-2)#