Question

How do you write the partial fraction decomposition of the rational expression `6/(x^2-25)^2`?

Answer 1

`6/(x^2-25)^2`

`= -3/(250(x-5))+3/(50(x-5)^2)+3/(250(x+5))+3/(50(x+5)^2)`

Explanation:

Note that:

`(x^2-25)^2 = ((x-5)(x+5))^2`

So need to solve:

`6/(x^2-25)^2 = A/(x-5)+B/(x-5)^2+C/(x+5)+D/(x+5)^2`

`=(Ax+(B-5A))/(x-5)^2 + (Cx+(D+5C))/(x+5)^2`

`=((Ax+(B-5A))(x+5)^2+(Cx+(D+5C))(x-5)^2)/(x^2-25)^2`

`=((Ax+(B-5A))(x^2+10x+25)+(Cx+(D+5C))(x^2-10x+25))/(x^2-25)^2`

`=((A+C)x^3+(B+D+5A-5C)x^2+(10B-10D-25A-25C)x+25(B+D-5A+5C))/(x^2-25)^2`

Hence:

`A+C=0`

`B+D+5A-5C=0`

`10B-10D-25A-25C=0`

`25(B+D-5A+5C) = 6`

From the first of these `C = -A`, so the rest become:

`B+D+10A=0`

`10(B-D)=0`

`25(B+D-10A) = 6`

From the second of these `D=B`, so the rest become:

`2B+10A=0`

`25(2B-10A) = 6`

From the first of these `B=-5A`, so the second becomes:

`25(-20A) = 6`

Hence `A=-3/250`, `B=3/50`, `C=3/250`, `D=3/50`

So:

`6/(x^2-25)^2`

`= -3/(250(x-5))+3/(50(x-5)^2)+3/(250(x+5))+3/(50(x+5)^2)`