How do you write the partial fraction decomposition of the rational expression #(5x-1)/(x^2-x-2)#?

1 Answer
Jan 4, 2016

#(5x-1)/(x^2-x-2)=3/(x-2)+2/(x+1)#

Explanation:

General form to break into partial fractions is
#(ax+b)/(cx^2+dx+e)=A/(x+C)+B/(x+D)#

#(5x-1)/(x^2-x-2)=(5x-1)/((x-2)(x+1))#

#(5x-1)/((x-2)(x+1))=A/(x-2)+B/(x+1)=(A(x+1)+B(x-2))/((x-2)(x+1))#

#(5x-1)/((x-2)(x+1))=(Ax+A+Bx-2B)/((x-2)(x+1))#

we can equate the coefficients of numerators of both sides
#A+B=5,A-2B=-1#
Subtracting both gives #3B=6#
or B = 2
And A=3
thus

#(5x-1)/(x^2-x-2)=3/(x-2)+2/(x+1)#