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

##### 1 Answer

#= -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

#B+D+10A=0#

#10(B-D)=0#

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

From the second of these

#2B+10A=0#

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

From the first of these

#25(-20A) = 6#

Hence

So:

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

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