How do you write the partial fraction decomposition of the rational expression # (x^3-x^2+1) / (x^4-x^3)#?

1 Answer
Dec 23, 2015

#(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1#

Explanation:

#(x^3-x^2+1) / (x^4-x^3) = (x^3-x^2)/ (x^4-x^3) + 1/(x^4-x^3)#

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

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

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

now focus on #1/(x^3(x-1))#

#1/(x^3(x-1)) = A/x^3+B/x^2+C/x+D/(x-1)#

Multiply both side by #x^3(x-1)#

#1 = A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3#

#1=Ax-A+Bx^2-Bx+Cx^3-Cx^2+Dx^3#

#1=x^3(C+D)+x^2(B-C)+x(A-B)-A#

#C+D = 0#
#B-C = 0#
#A-B = 0#
#A = -1#

Just by looking we have

#A = - 1#
#B = -1#
#C = -1#
#D = 1#

So

#(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1)#