1. **State the problem:** Simplify the expression $$\frac{x-4}{x^2+8x+15} \div \frac{x^2-16}{2x^2+7x+3}$$ and state the restrictions on $x$. 2. **Rewrite the division as multiplication by the reciprocal:** $$\frac{x-4}{x^2+8x+15} \times \frac{2x^2+7x+3}{x^2-16}$$ 3. **Factor all polynomials:** - $x^2+8x+15 = (x+3)(x+5)$ - $x^2-16 = (x-4)(x+4)$ - $2x^2+7x+3 = (2x+1)(x+3)$ So the expression becomes: $$\frac{x-4}{(x+3)(x+5)} \times \frac{(2x+1)(x+3)}{(x-4)(x+4)}$$ 4. **Combine into a single fraction:** $$\frac{(x-4)(2x+1)(x+3)}{(x+3)(x+5)(x-4)(x+4)}$$ 5. **Cancel common factors:** - Cancel $(x-4)$ numerator and denominator - Cancel $(x+3)$ numerator and denominator Intermediate step showing cancellation: $$\frac{\cancel{x-4}(2x+1)\cancel{(x+3)}}{\cancel{(x+3)}(x+5)\cancel{(x-4)}(x+4)} = \frac{2x+1}{(x+5)(x+4)}$$ 6. **Final simplified expression:** $$\frac{2x+1}{(x+5)(x+4)}$$ 7. **State restrictions:** - Denominators cannot be zero. - From original denominators: - $x^2+8x+15 = 0 \Rightarrow (x+3)(x+5)=0 \Rightarrow x \neq -3, -5$ - $x^2-16=0 \Rightarrow (x-4)(x+4)=0 \Rightarrow x \neq 4, -4$ - $2x^2+7x+3=0 \Rightarrow (2x+1)(x+3)=0 \Rightarrow x \neq -\frac{1}{2}, -3$ Combining all restrictions: $$x \neq -3, -5, 4, -4, -\frac{1}{2}$$