How do you write the partial fraction decomposition of the rational expression (3x^2+2x)/(x^2-4)3x2+2xx24?

1 Answer
Nov 26, 2016

The answer is =3-2/(x+2)+4/(x-2)=32x+2+4x2

Explanation:

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

Let's do a long division

color(white)(aaaaa)aaaaa3x^2+2x3x2+2xcolor(white)(aaaaa)aaaaax^2-4x24

color(white)(aaaaa)aaaaa3x^2-123x212color(white)(aaaaa)aaaaa33

color(white)(aaaaaa)aaaaaa0+2x+120+2x+12

(3x^2+2x)/(x^2-4)=3+(2x+12)/(x^2-4)3x2+2xx24=3+2x+12x24

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)2x+12x24=2x+12(x+2)(x2)=Ax+2+Bx2

=(A(x-2)+B(x+2))/((x+2)(x-2))=A(x2)+B(x+2)(x+2)(x2)

So, (2x+12)=(A(x-2)+B(x+2))(2x+12)=(A(x2)+B(x+2))

Let x=2x=2, 16=4B16=4B, =>, B=4B=4

Let x=-2x=2, 8=-4A8=4A, =>, A=-2A=2

So, (2x+12)/(x^2-4)=-2/(x+2)+4/(x-2)2x+12x24=2x+2+4x2

And finally we have

(3x^2+2x)/(x^2-4)=3-2/(x+2)+4/(x-2)3x2+2xx24=32x+2+4x2