partial fractions
posted by gabs .
(7x^216)/(x^22x)
solve by using partial fraction decomposition. i cant figure out how to do this because after i split up the bottom the A and B don't end up having an x^2 in order to equal the numerator of the original fraction. so confused!

partial fractions 
Count Iblis
The problem here is that the partial expansion looks like:
A + B/(x2) + C/x 
partial fractions 
Reiny
To avoid the problem of the x^2 in the top, let's do a division first.
By long division ....
(7x^  16)/(x^2  2x)
= 7 + (14x16)/(x^2  2x)
so now let's just work on that last term.
let (14x 16)/(x^  2x) = A/x + B/(x2)
= (A(x2) + Bx) / (x(x2))
then 14x  16 = A(x2) + Bx
let x = 0
16 = 2A
A = 8
let x = 2
12 = 2B
B = 6
so (7x^216)/(x^22x) = 7 + 8/x + 6/(x2) 
partial fractions 
Count Iblis
R(x) = (7x^216)/(x^22x) =
(7 x^2  16)/[x(x2)]
Near x = 0, the singular behavior is:
R(x) = 1/x (7 x^2  16)/(x2) =
1/x (16/(2)) + nonsingular terms =
8/x + nonsingular terms.
Near x = 2, the singular behavior is:
R(x) = 1/(x2) (7 x^2  16)/x =
1/(x2) [7*416]/2 + nonsingular terms
=
6/(x2) + nonsingular terms.
R(x) minus alll singular terms =
R(x)  8/x  6/(x2)
Now a rational function from which we have subtracted all singular terms doesn't ave any singularities, so it is actually a polynomial. We can find this polynomial by considering the behavior of the function at infinity. We see that the limit of x to infinity of R(x) exists and is equal to 7. The singular terms all go to zero at infinity. So,
R(x) minus the singular terms, which we know to be polynomial tends to zero at infinity, therefore the polynomial is a
cosntant function equal to 7 everwhere.
We can thus conclude that:
R(x)  8/x  6/(x2) = 7 >
R(x) = 7 + 8/x + 6/(x2)
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