Divide and state the quotient in simplest for \( \frac{x^{2}-4 x-45}{x+6} \div \frac{x^{2}-3 x-40}{x+6} \) \( \frac{9}{8} \) \( \frac{x-8}{x-9} \) \( \frac{x-9}{x-8} \) \( \frac{x^{2}-4 x-45}{x^{2}-3 x-40} \)

To simplify the expression \[ \frac{x^{2}-4x-45}{x+6} \div \frac{x^{2}-3x-40}{x+6}, \] we can rewrite the division as multiplication by the reciprocal: \[ \frac{x^{2}-4x-45}{x+6} \cdot \frac{x+6}{x^{2}-3x-40}. \] The \(x+6\) terms in the numerator and denominator cancel out, provided \(x \neq -6\): \[ \frac{x^{2}-4x-45}{x^{2}-3x-40}. \] Next, we need to factor both the numerator and the denominator. 1. **Factoring the numerator \(x^{2}-4x-45\)**: - We look for two numbers that multiply to \(-45\) and add to \(-4\). These numbers are \(-9\) and \(5\). - Thus, we can factor it as: \[ x^{2}-4x-45 = (x-9)(x+5). \] 2. **Factoring the denominator \(x^{2}-3x-40\)**: - We look for two numbers that multiply to \(-40\) and add to \(-3\). These numbers are \(-8\) and \(5\). - Thus, we can factor it as: \[ x^{2}-3x-40 = (x-8)(x+5). \] Now we can substitute the factored forms back into our expression: \[ \frac{(x-9)(x+5)}{(x-8)(x+5)}. \] Again, the \(x+5\) terms in the numerator and denominator cancel out, provided \(x \neq -5\): \[ \frac{x-9}{x-8}. \] Thus, the quotient in simplest form is: \[ \frac{x-9}{x-8}. \] Therefore, the answer is \(\frac{x-9}{x-8}\).