Divide and simplify \[ \frac{x^{2}-5 x-24}{2 x^{2}-128} \div \frac{x^{2}+11 x+24}{x^{2}+16 x+64} \]

To divide and simplify the expression \[ \frac{x^{2}-5 x-24}{2 x^{2}-128} \div \frac{x^{2}+11 x+24}{x^{2}+16 x+64}, \] we start by rewriting it as multiplication by the reciprocal: \[ \frac{x^{2}-5 x-24}{2 x^{2}-128} \cdot \frac{x^{2}+16 x+64}{x^{2}+11 x+24}. \] Next, we simplify each polynomial. 1. **Factoring the polynomials:** - For \(x^{2}-5x-24\): \[ x^{2}-5x-24 = (x - 8)(x + 3). \] - For \(2x^{2}-128\): \[ 2x^{2}-128 = 2(x^{2}-64) = 2(x-8)(x+8). \] - For \(x^{2}+11x+24\): \[ x^{2}+11x+24 = (x + 3)(x + 8). \] - For \(x^{2}+16x+64\): \[ x^{2}+16x+64 = (x + 8)^2. \] 2. **Substituting back into the expression:** Now we rewrite our expression with the factored forms: \[ \frac{(x-8)(x+3)}{2(x-8)(x+8)} \cdot \frac{(x+8)^2}{(x+3)(x+8)}. \] 3. **Canceling common factors:** - We can cancel \((x - 8)\) from the numerator and the denominator. - We can cancel one \((x + 8)\) from the numerator and the denominator. - We can also cancel \((x + 3)\) from the numerator and denominator. After canceling, we have: \[ \frac{1}{2} \cdot \frac{x + 8}{1} = \frac{x + 8}{2}. \] Thus, the final simplified result is: \[ \boxed{\frac{x + 8}{2}}. \]