Express $\frac{2x^3 - 4x^2 - 5x + 3}{(x-3)(x^2-9)}$ in partial fractions.

1. Expand the denominator and perform long division since the degree of the numerator equals the degree of the denominator.

Denominator: $(x-3)(x^2-9) = (x-3)(x-3)(x+3) = x^3 - 3x^2 - 9x + 27$

Since the numerator $2x^3 - 4x^2 - 5x + 3$ has degree 3, we divide it by the denominator:
$\frac{2x^3 - 4x^2 - 5x + 3}{x^3 - 3x^2 - 9x + 27} = 2 + \frac{(2x^3 - 4x^2 - 5x + 3) - 2(x^3 - 3x^2 - 9x + 27)}{(x-3)^2(x+3)}$
$= 2 + \frac{2x^2 + 13x - 51}{(x-3)^2(x+3)}$

2. Set up the partial fraction decomposition for the proper fraction part.

Let $\frac{2x^2 + 13x - 51}{(x-3)^2(x+3)} = \frac{A}{x-3} + \frac{B}{(x-3)^2} + \frac{C}{x+3}$

Multiplying by $(x-3)^2(x+3)$ gives:
$2x^2 + 13x - 51 = A(x-3)(x+3) + B(x+3) + C(x-3)^2$

3. Solve for the constants $A$, $B$, and $C$.

Substitute $x = 3$:
$2(3)^2 + 13(3) - 51 = B(3+3)$
$18 + 39 - 51 = 6B \implies 6 = 6B \implies B = 1$

Substitute $x = -3$:
$2(-3)^2 + 13(-3) - 51 = C(-3-3)^2$
$18 - 39 - 51 = 36C \implies -72 = 36C \implies C = -2$

Compare coefficients of $x^2$:
$2 = A + C$
$2 = A - 2 \implies A = 4$

4. Write the final expression in partial fractions.

$\frac{2x^3 - 4x^2 - 5x + 3}{(x-3)(x^2-9)} = 2 + \frac{4}{x-3} + \frac{1}{(x-3)^2} - \frac{2}{x+3}$