# partial fractions

posted by gabs

(7x^2-16)/(x^2-2x)

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!

1. Count Iblis

The problem here is that the partial expansion looks like:

A + B/(x-2) + C/x

2. 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 + (14x-16)/(x^2 - 2x)

so now let's just work on that last term.

let (14x -16)/(x^ - 2x) = A/x + B/(x-2)
= (A(x-2) + Bx) / (x(x-2))

then 14x - 16 = A(x-2) + Bx
let x = 0
-16 = -2A
A = 8

let x = 2
12 = 2B
B = 6

so (7x^2-16)/(x^2-2x) = 7 + 8/x + 6/(x-2)

3. Count Iblis

R(x) = (7x^2-16)/(x^2-2x) =

(7 x^2 - 16)/[x(x-2)]

Near x = 0, the singular behavior is:

R(x) = 1/x (7 x^2 - 16)/(x-2) =

1/x (-16/(-2)) + non-singular terms =

8/x + non-singular terms.

Near x = 2, the singular behavior is:

R(x) = 1/(x-2) (7 x^2 - 16)/x =

1/(x-2) [7*4-16]/2 + non-singular terms

=

6/(x-2) + non-singular terms.

R(x) minus alll singular terms =

R(x) - 8/x - 6/(x-2)

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/(x-2) = 7 -------->

R(x) = 7 + 8/x + 6/(x-2)

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