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

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Answer 1 · 2018-03-08T12:42:15

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

We need to write these in terms of each factors.

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

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

Putting in #x=-2#:
#(-2)^2=A(-2+2)+B(-2-1)#
#4=-3B#
#B=-4/3#

Putting in #x=1#:
#1^2=A(1+2)+B(1-1)#
#1=3A#
#A=1/3#

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

Answer 2 · 2018-03-08T12:53:37

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

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

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

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

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

Now, I decomposed fraction into basic ones,

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

After expanding denominator,

#A*(x+2)+B*(x-1)=x-2#

Set #x=-2#, #-3B=-4#, so #B=4/3#

Set #x=1#, #3A=-1#, so #A=-1/3#

Hence,

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

Thus,

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

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

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