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

Question

Serenity Johnson · Accepted Answer

#(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)#

Lincoln Anderson · Answer

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]

Lincoln Anderson · Answer

undefined To decompose the rational expression (4x + 4)/(x^2(x + 2)), follow these steps:

1. Factor the denominator: x^2(x + 2) = x^2 * (x + 2).
2. Write the expression as a sum of fractions with simpler denominators.
3. The decomposition will have the form: A/x + B/x^2 + C/(x + 2), where A, B, and C are constants to be determined.
4. Multiply both sides of the equation by the common denominator x^2(x + 2).
5. Simplify and solve for A, B, and C.

By solving the equation, you'll find the values of A, B, and C.