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

To simplify the expression \[ \frac{3x+3}{x-2} \div \frac{x^2-2x-3}{x^2-4}, \] we first rewrite the division as multiplication by the reciprocal: \[ \frac{3x+3}{x-2} \times \frac{x^2-4}{x^2-2x-3}. \] Next, we can factor the components. The numerator \(3x+3\) can be factored as \(3(x+1)\). For \(x^2-2x-3\), we look for two numbers that multiply to \(-3\) and add to \(-2\), which are \(-3\) and \(1\). Thus, it factors to \((x-3)(x+1)\). The expression \(x^2-4\) can be factored as \((x-2)(x+2)\). Now, substituting the factored forms, we get: \[ \frac{3(x+1)}{x-2} \times \frac{(x-2)(x+2)}{(x-3)(x+1)}. \] Now we can cancel out the common factors. The \(x+1\) in the numerator and denominator cancel, and \(x-2\) in the numerator and denominator also cancel: \[ \frac{3 \cancel{(x+1)}}{\cancel{(x-2)}} \times \frac{\cancel{(x-2)}(x+2)}{(x-3)\cancel{(x+1)}} = \frac{3(x+2)}{x-3}. \] Therefore, the simplified form of the original expression is: \[ \frac{3(x+2)}{x-3}. \] You can express it as: \[ \frac{3x + 6}{x - 3}. \]