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)aaaaa∣∣x^2-4x2−4
color(white)(aaaaa)aaaaa3x^2-123x2−12color(white)(aaaaa)aaaaa∣∣33
color(white)(aaaaaa)aaaaaa0+2x+120+2x+12
(3x^2+2x)/(x^2-4)=3+(2x+12)/(x^2-4)3x2+2xx2−4=3+2x+12x2−4
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)2x+12x2−4=2x+12(x+2)(x−2)=Ax+2+Bx−2
=(A(x-2)+B(x+2))/((x+2)(x-2))=A(x−2)+B(x+2)(x+2)(x−2)
So, (2x+12)=(A(x-2)+B(x+2))(2x+12)=(A(x−2)+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+12x2−4=−2x+2+4x−2
And finally we have
(3x^2+2x)/(x^2-4)=3-2/(x+2)+4/(x-2)3x2+2xx2−4=3−2x+2+4x−2