1. **State the problem:** Simplify the expression $ \frac{x^2 + 4x - 32}{2x^2 + 9x - 5} \cdot \frac{3x^2 - 75}{3x^2 - 11x - 4} \div \frac{6x^2 - 18x - 60}{x^3 - 4x} $. 2. **Recall the formula and rules:** Division of fractions means multiplying by the reciprocal. So, $$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C}.$$ 3. **Factor all polynomials:** - $x^2 + 4x - 32 = (x + 8)(x - 4)$ - $2x^2 + 9x - 5 = (2x - 1)(x + 5)$ - $3x^2 - 75 = 3(x^2 - 25) = 3(x - 5)(x + 5)$ - $3x^2 - 11x - 4 = (3x + 1)(x - 4)$ - $6x^2 - 18x - 60 = 6(x^2 - 3x - 10) = 6(x - 5)(x + 2)$ - $x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)$ 4. **Rewrite the expression with factors:** $$\frac{(x + 8)(x - 4)}{(2x - 1)(x + 5)} \times \frac{3(x - 5)(x + 5)}{(3x + 1)(x - 4)} \times \frac{x(x - 2)(x + 2)}{6(x - 5)(x + 2)}$$ 5. **Cancel common factors:** - $x - 4$ cancels - $x + 5$ cancels - $x - 5$ cancels - $x + 2$ cancels Intermediate step showing cancellation: $$\frac{(x + 8)\cancel{(x - 4)}}{(2x - 1)\cancel{(x + 5)}} \times \frac{3\cancel{(x - 5)}\cancel{(x + 5)}}{(3x + 1)\cancel{(x - 4)}} \times \frac{x(x - 2)\cancel{(x + 2)}}{6\cancel{(x - 5)}\cancel{(x + 2)}}$$ 6. **Multiply remaining factors:** $$\frac{(x + 8)}{(2x - 1)} \times \frac{3}{(3x + 1)} \times \frac{x(x - 2)}{6} = \frac{3x(x - 2)(x + 8)}{6(2x - 1)(3x + 1)}$$ 7. **Simplify the coefficient:** $$\frac{3}{6} = \frac{1}{2}$$ 8. **Final simplified expression:** $$\boxed{\frac{x(x - 2)(x + 8)}{2(2x - 1)(3x + 1)}}$$