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

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Answer 1 · 2016-01-06T09:15:50

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

#(x+2)/(x(x-4))#

#(x+2)/(x(x-4)) = A/x + B/(x-4)#

#(x+2)/(x(x-4)) =(A(x-4)+B(x))/(x(x-4))#

#(x+2) = A(x-4)+B(x)#

Let #x=0# This done to remove the #B#
#(0+2) = A(0-4)+B(0)#

#2=-4A#

#2/-4 =(-4A)/-4#
#-1/2 = A#

#A=-1/2#

Now let #x=4# Which makes #(x-4)# as zero and thus eliminating #A# and we can solve for #B#

#(4+2)=A(4-4)+B(4)#
#6=4B#
#6/4 = (4B)/4#
#3/2 = B#

#B=3/2#

Therefore,

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

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