1. **State the problem:** Simplify the expression $$\frac{x^2 - 49}{x^2 + 2x - 35} \times \frac{2x^3 - 18x^2 + 40x}{3x^2 - 21x}$$ and find the restrictions on the variable $x$. 2. **Factor all polynomials:** - Numerator 1: $x^2 - 49 = (x - 7)(x + 7)$ (difference of squares). - Denominator 1: $x^2 + 2x - 35 = (x + 7)(x - 5)$. - Numerator 2: $2x^3 - 18x^2 + 40x = 2x(x^2 - 9x + 20)$. Factor quadratic: $x^2 - 9x + 20 = (x - 5)(x - 4)$. So numerator 2: $2x(x - 5)(x - 4)$. - Denominator 2: $3x^2 - 21x = 3x(x - 7)$. 3. **Rewrite the expression with factors:** $$\frac{(x - 7)(x + 7)}{(x + 7)(x - 5)} \times \frac{2x(x - 5)(x - 4)}{3x(x - 7)}$$ 4. **Cancel common factors:** - Cancel $(x + 7)$ from numerator 1 and denominator 1. - Cancel $(x - 5)$ from numerator 2 and denominator 1. - Cancel $(x - 7)$ from numerator 1 and denominator 2. - Cancel $x$ from numerator 2 and denominator 2. Intermediate step showing cancellation: $$\frac{\cancel{(x - 7)}\cancel{(x + 7)}}{\cancel{(x + 7)}\cancel{(x - 5)}} \times \frac{2\cancel{x}\cancel{(x - 5)}(x - 4)}{3\cancel{x}\cancel{(x - 7)}} = \frac{1}{1} \times \frac{2(x - 4)}{3} = \frac{2(x - 4)}{3}$$ 5. **Simplified expression:** $$\frac{2(x - 4)}{3}$$ 6. **State restrictions:** - Denominators cannot be zero. - From original denominators: - $x + 7 \neq 0 \Rightarrow x \neq -7$ - $x - 5 \neq 0 \Rightarrow x \neq 5$ - $3x(x - 7) \neq 0 \Rightarrow x \neq 0, x \neq 7$ **Final answer:** $$\boxed{\frac{2(x - 4)}{3}, \quad x \neq -7, 0, 5, 7}$$