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

2 Answers
Mar 8, 2018

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

Explanation:

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))

Mar 8, 2018

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

Explanation:

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)