PreCalc
posted by Micki on .
How do I solve this via partial fraction decomposition?
(x^2+9)/(x^42x^28)

The bottom factors to (x^24)(x^2+2)
= (x+2)x2)x^2+4)
so let
(x^2+9)/(x^42x^28) = A/(x+2) + B/(x2) + C/(x^2+2)
( A(x2)(x^2+2) + B(x+2)(x^2+2) + C(x+2)(x2) ) = x^2 + 9
since the denominators on the left and right are the same.
This must be true for all values of x
let x=2 > 24B = 13 or B = 13/24
let x=2 >  24A = 13 or A = 13/24
let x = 1 > 3A + 9B  3C = 10
sub in the values of A and B to get
C = 7/6
so (x^2+9)/(x^42x^28)
= 13/(24(x+2)) + 13/(24(x2))  7/(6(x^2+2)) 
Thank you. But what about the "Cx+D" part?

You are correct, Micki, that there is supposed to be a (Cx+D) over a quadratic denominator. Luckily, in this case, C=0, so it worked out. Had it been otherwise, I'm sure it would have become apparent.