Can (-12x^2 + 33x - 6)/(x^3 - 4x) be presented as a sum of elementary fractions?

We have to write (-12x^2 + 33x - 6)/(x^3 - 4x) as the sum
of elementary fractions.

(-12x^2 + 33x - 6)/(x^3 -
4x)

=> (-12x^2 + 33x - 6)/x(x^2 -
4)

=> (-12x^2 + 33x - 6)/x(x - 2)(x +
2)

Let this be equal to A/x + B/(x - 2) + C/(x +
2)

=> [A(x^2 - 4) + B(x^2 + 2x) + C(x^2 - 2x)]/x(x^2
- 4) = (-12x^2 + 33x - 6)/(x^3 - 4x)

=> A(x^2 - 4) +
B(x^2 + 2x) + C(x^2 - 2x) = (-12x^2 + 33x - 6)

=>
Ax^2 - 4A + Bx^2 + 2Bx + Cx^2 - 2Cx = -12x^2 + 33x -
6

=> A + B + C = -12, 2B - 2C = 33 and -4A =
-6

=> A = 6/4 = 3/2

B +
C = -13.5 and B - C = 16.5

Add the two 2B =
3

B = 1.5

C =
-15

The given fraction can be written as
1.5/x + 1.5/(x - 2) - 15/(x + 2)