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

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Answer 1 · 2018-03-22T12:18:45

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

As we have #x^2(x+2)# in the denominator, we can have partial fractions as

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

or #4x+4=Ax(x+2)+B(x+2)+Cx^2# ..............(P)

If we put #x=0# in (P), we get #2B=4# i.e. #B=2#

and if we put #x=-2#, then #4C=-4# or #C=-1#

Comparing coefficient of #x^2# in (P), we get

#A+C=0# i.e. #A=-C=1#

Hence, partial fractions are

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

Answer 2 · 2018-03-22T12:22:46

Partial fraction : # (4x+4)/(x^2(x+2))= 1/x+2/x^2-1/(x+2)#

Let # (4x+4)/(x^2(x+2))= A/x+B/x^2+C/(x+2)# or

#(4x+4)/(x^2(x+2))=(Ax(x+2)+B(x+2)+Cx^2)/(x^2(x+2))# or

#Ax(x+2)+B(x+2)+Cx^2=4x+4# or

# x^2(A+C)+x(2A+B)+2B=4x+4#

Equating the co-efficient of #x^2# in both sides we get, #A+C=0#

Equating the co-efficient of #x# in both sides we get, #2A+B=4#

Equating constant term in both sides we get, #2B=4or B=2#

#B=2 :. 2A+2=4 :. 2A = 2 :. A=1; A+C=0#

# :. C=-A :. C= -1# Hence partial fraction is

# (4x+4)/(x^2(x+2))= 1/x+2/x^2-1/(x+2)# [Ans]