Multiply. \[ \frac{4 x-8}{x-3} \cdot \frac{x+3}{x^{2}-4} \] Simplify your answer as much as possible. \( \square \frac{\square}{\square} \)

To simplify the expression \[ \frac{4 x-8}{x-3} \cdot \frac{x+3}{x^{2}-4}, \] we first factor each component where possible. 1. The numerator \(4x - 8\) can be factored as \(4(x - 2)\). 2. The denominator \(x^2 - 4\) can be factored since it is a difference of squares as \((x - 2)(x + 2)\). Now substituting these factorizations into the expression gives us: \[ \frac{4(x - 2)}{x - 3} \cdot \frac{x + 3}{(x - 2)(x + 2)}. \] Next, we can cancel the common factor of \((x - 2)\) in the numerator and denominator (as long as \(x \neq 2\)): \[ = \frac{4}{x - 3} \cdot \frac{x + 3}{x + 2}. \] Now we multiply the remaining fractions: \[ = \frac{4(x + 3)}{(x - 3)(x + 2)}. \] Thus, the simplified expression is: \[ \frac{4(x + 3)}{(x - 3)(x + 2)}. \] In simplified form: \[ \boxed{\frac{4(x+3)}{(x-3)(x+2)}} \]