1. **State the problem:** Multiply the rational expressions $$\frac{9y^2-4}{4y-12} \times \frac{18-6y}{9y^2+12y+4}$$. 2. **Factor all polynomials:** - Numerator of first fraction: $$9y^2-4 = (3y)^2 - 2^2 = (3y-2)(3y+2)$$ (difference of squares). - Denominator of first fraction: $$4y-12 = 4(y-3)$$ (factor out 4). - Numerator of second fraction: $$18-6y = 6(3 - y) = 6(-(y-3)) = -6(y-3)$$ (factor out 6 and rewrite). - Denominator of second fraction: $$9y^2 + 12y + 4 = (3y + 2)^2$$ (perfect square trinomial). 3. **Rewrite the expression with factors:** $$\frac{(3y-2)(3y+2)}{4(y-3)} \times \frac{-6(y-3)}{(3y+2)^2}$$ 4. **Multiply the numerators and denominators:** $$\frac{(3y-2)(3y+2) \times -6(y-3)}{4(y-3) \times (3y+2)^2}$$ 5. **Cancel common factors:** - Cancel $(y-3)$ in numerator and denominator: $$\frac{(3y-2)(3y+2) \times -6\cancel{(y-3)}}{4\cancel{(y-3)} \times (3y+2)^2}$$ - Cancel one $(3y+2)$ factor: $$\frac{(3y-2)\cancel{(3y+2)} \times -6}{4 \times (3y+2)\cancel{(3y+2)}}$$ 6. **Simplify constants:** $$\frac{-6(3y-2)}{4(3y+2)}$$ 7. **Simplify the fraction $$\frac{-6}{4}$$ by dividing numerator and denominator by 2:** $$\frac{\cancel{-6}^{-3}}{\cancel{4}^2} = \frac{-3}{2}$$ 8. **Final simplified expression:** $$\frac{-3(3y-2)}{2(3y+2)} = \frac{-9y + 6}{2(3y+2)}$$ **Answer:** $$\boxed{\frac{-3(3y-2)}{2(3y+2)}}$$