How do you write the partial fraction decomposition of the rational expression 6/(x^2-25)^26(x2−25)2?

1 Answer
George C.
Dec 30, 2015

6/(x^2-25)^26(x2−25)2

= -3/(250(x-5))+3/(50(x-5)^2)+3/(250(x+5))+3/(50(x+5)^2)=−3250(x−5)+350(x−5)2+3250(x+5)+350(x+5)2

Explanation:

Note that:

(x^2-25)^2 = ((x-5)(x+5))^2(x2−25)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)^26(x2−25)2=Ax−5+B(x−5)2+Cx+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

=((Ax+(B-5A))(x+5)^2+(Cx+(D+5C))(x-5)^2)/(x^2-25)^2=(Ax+(B−5A))(x+5)2+(Cx+(D+5C))(x−5)2(x2−25)2

=((Ax+(B-5A))(x^2+10x+25)+(Cx+(D+5C))(x^2-10x+25))/(x^2-25)^2=(Ax+(B−5A))(x2+10x+25)+(Cx+(D+5C))(x2−10x+25)(x2−25)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=(A+C)x3+(B+D+5A−5C)x2+(10B−10D−25A−25C)x+25(B+D−5A+5C)(x2−25)2

Hence:

A+C=0A+C=0

B+D+5A-5C=0B+D+5A−5C=0

10B-10D-25A-25C=010B−10D−25A−25C=0

25(B+D-5A+5C) = 625(B+D−5A+5C)=6

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

B+D+10A=0B+D+10A=0

10(B-D)=010(B−D)=0

25(B+D-10A) = 625(B+D−10A)=6

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

2B+10A=02B+10A=0

25(2B-10A) = 625(2B−10A)=6

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

25(-20A) = 625(−20A)=6

Hence A=-3/250A=−3250, B=3/50B=350, C=3/250C=3250, D=3/50D=350

So:

6/(x^2-25)^26(x2−25)2

= -3/(250(x-5))+3/(50(x-5)^2)+3/(250(x+5))+3/(50(x+5)^2)=−3250(x−5)+350(x−5)2+3250(x+5)+350(x+5)2

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