Question

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

Answer 1

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

Explanation:

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)`

Answer 2

Partial fraction : `(4x+4)/(x^2(x+2))= 1/x+2/x^2-1/(x+2)`

Explanation:

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]