How do you express # (x+2) / (x(x-4))# in partial fractions?

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Answer 1 · 2018-05-19T18:29:47

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

Here,

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

#color(white)((x+2)/(x(x-4)))=1/2[(3x-(x-4))/(x(x-4))]#

#color(white)((x+2)/(x(x-4)))=1/2[(3x)/(x(x-4))-(x-4)/(x(x-4))]#

#color(white)((x+2)/(x(x-4)))=1/2[3/(x-4)-1/x]#

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

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

#=>x+2=Ax+B(x-4)#

Take ,

#x=4=>4+2=A(4)=>4A=6=>A=3/2#

#x=0=>0+2=B(-4)=>B=-2/4=-1/2#

Subst. values of #A and B# into #(2)#

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

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