The problem asks us to resolve the given rational function into partial fractions. The function is $\frac{2x^2 - 18x + 12}{x^3 - 4x}$.

1. Problem Description

The problem asks us to resolve the given rational function into partial fractions. The function is $\frac{2x^2 - 18x + 12}{x^3 - 4x}$.

2. Solution Steps

First, we factor the denominator:

$x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2)$.

Thus, we can express the rational function as:

$\frac{2x^2 - 18x + 12}{x(x-2)(x+2)} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+2}$.

Multiplying both sides by $x(x-2)(x+2)$ gives:

$2x^2 - 18x + 12 = A(x-2)(x+2) + Bx(x+2) + Cx(x-2)$.

$2x^2 - 18x + 12 = A(x^2 - 4) + B(x^2 + 2x) + C(x^2 - 2x)$.

$2x^2 - 18x + 12 = Ax^2 - 4A + Bx^2 + 2Bx + Cx^2 - 2Cx$.

Grouping like terms, we get:

$2x^2 - 18x + 12 = (A+B+C)x^2 + (2B-2C)x - 4A$.

By comparing coefficients, we obtain the following system of equations:

$A + B + C = 2$

$2B - 2C = -18$

$-4A = 12$

From the third equation, we get $A = -3$.

Substituting $A = -3$ into the first equation, we get $-3 + B + C = 2$, so $B + C = 5$.

Dividing the second equation by 2, we get $B - C = -9$.

Adding the equations $B + C = 5$ and $B - C = -9$ gives $2B = -4$, so $B = -2$.

Since $B + C = 5$, we have $-2 + C = 5$, so $C = 7$.

Therefore, $A = -3$, $B = -2$, and $C = 7$.

Substituting these values back into the partial fraction decomposition, we have:

$\frac{2x^2 - 18x + 12}{x(x-2)(x+2)} = \frac{-3}{x} + \frac{-2}{x-2} + \frac{7}{x+2}$.

So, the partial fraction decomposition is $-\frac{3}{x} - \frac{2}{x-2} + \frac{7}{x+2}$.

3. Final Answer

$-\frac{3}{x} - \frac{2}{x-2} + \frac{7}{x+2}$