1. **State the problem:** Simplify the expression $$\left(\frac{x^2-4x-5}{x^2-3x-10}\right)\left(\frac{9x+3x^2}{x^2-9}\right)$$ and state the restrictions on $x$. 2. **Factor all polynomials:** - Numerator of first fraction: $$x^2-4x-5 = (x-5)(x+1)$$ - Denominator of first fraction: $$x^2-3x-10 = (x-5)(x+2)$$ - Numerator of second fraction: $$9x+3x^2 = 3x(3+x) = 3x(x+3)$$ - Denominator of second fraction: $$x^2-9 = (x-3)(x+3)$$ 3. **Rewrite the expression with factored forms:** $$\left(\frac{(x-5)(x+1)}{(x-5)(x+2)}\right)\left(\frac{3x(x+3)}{(x-3)(x+3)}\right)$$ 4. **Cancel common factors:** - Cancel $(x-5)$ from numerator and denominator: $$\frac{\cancel{(x-5)}(x+1)}{\cancel{(x-5)}(x+2)} = \frac{x+1}{x+2}$$ - Cancel $(x+3)$ from numerator and denominator: $$\frac{3x\cancel{(x+3)}}{(x-3)\cancel{(x+3)}} = \frac{3x}{x-3}$$ 5. **Multiply the simplified fractions:** $$\frac{x+1}{x+2} \times \frac{3x}{x-3} = \frac{3x(x+1)}{(x+2)(x-3)}$$ 6. **State restrictions:** - Denominators cannot be zero: - From $x^2-3x-10 = (x-5)(x+2)$, $x \neq 5$, $x \neq -2$ - From $x^2-9 = (x-3)(x+3)$, $x \neq 3$, $x \neq -3$ **Final answer:** $$\frac{3x(x+1)}{(x+2)(x-3)}$$ **Restrictions:** $x \neq 5, -2, 3, -3$