1. **Stating the problem:** Simplify the product of the two rational expressions: $$\frac{a^2 + 4a + 4}{2 - a} \cdot \frac{2a^2 - 8a + 8}{a^2 - 4}$$ 2. **Recall important formulas and rules:** - Factor quadratic expressions when possible. - Remember difference of squares: $$x^2 - y^2 = (x - y)(x + y)$$ - When multiplying fractions, multiply numerators together and denominators together. - Simplify by canceling common factors. 3. **Factor each polynomial:** - $$a^2 + 4a + 4 = (a + 2)^2$$ - $$2a^2 - 8a + 8 = 2(a^2 - 4a + 4) = 2(a - 2)^2$$ - $$a^2 - 4 = (a - 2)(a + 2)$$ 4. **Rewrite the expression with factored forms:** $$\frac{(a + 2)^2}{2 - a} \cdot \frac{2(a - 2)^2}{(a - 2)(a + 2)}$$ 5. **Simplify the denominator $$2 - a$$:** Note that $$2 - a = -(a - 2)$$, so: $$\frac{(a + 2)^2}{-(a - 2)} \cdot \frac{2(a - 2)^2}{(a - 2)(a + 2)}$$ 6. **Multiply numerators and denominators:** Numerator: $$(a + 2)^2 \cdot 2 (a - 2)^2 = 2 (a + 2)^2 (a - 2)^2$$ Denominator: $$-(a - 2) \cdot (a - 2)(a + 2) = - (a - 2)^2 (a + 2)$$ 7. **Write the fraction:** $$\frac{2 (a + 2)^2 (a - 2)^2}{- (a - 2)^2 (a + 2)}$$ 8. **Cancel common factors:** Cancel $$(a - 2)^2$$ from numerator and denominator: $$\frac{2 (a + 2)^2 \cancel{(a - 2)^2}}{- \cancel{(a - 2)^2} (a + 2)} = \frac{2 (a + 2)^2}{- (a + 2)}$$ 9. **Simplify further:** Cancel one $$(a + 2)$$: $$\frac{2 \cancel{(a + 2)} (a + 2)}{- \cancel{(a + 2)}} = \frac{2 (a + 2)}{-1} = -2 (a + 2)$$ 10. **Final answer:** $$\boxed{-2 (a + 2)}$$ This matches the expected result.