1. **State the problem:** Multiply and simplify the expression $$\frac{14a - 21a^2}{9a^2 + 6a + 1} \times \frac{9a^2 + 9a + 2}{9a^2 - 4}$$ 2. **Factor all polynomials where possible:** - Numerator of first fraction: $14a - 21a^2 = 7a(2 - 3a)$ - Denominator of first fraction: $9a^2 + 6a + 1 = (3a + 1)^2$ - Numerator of second fraction: $9a^2 + 9a + 2$ (try to factor) The quadratic factors as $(3a + 2)(3a + 1)$ because $3a \times 3a = 9a^2$, $2 \times 1 = 2$, and $3a \times 1 + 2 \times 3a = 3a + 6a = 9a$. - Denominator of second fraction: $9a^2 - 4 = (3a - 2)(3a + 2)$ (difference of squares) 3. **Rewrite the expression with factored forms:** $$\frac{7a(2 - 3a)}{(3a + 1)^2} \times \frac{(3a + 2)(3a + 1)}{(3a - 2)(3a + 2)}$$ 4. **Cancel common factors:** - $(3a + 2)$ appears in numerator and denominator, cancel: $$\frac{7a(2 - 3a)}{(3a + 1)^2} \times \frac{\cancel{(3a + 2)}(3a + 1)}{(3a - 2)\cancel{(3a + 2)}}$$ - $(3a + 1)$ appears squared in denominator and once in numerator, cancel one: $$\frac{7a(2 - 3a)}{\cancel{(3a + 1)}(3a + 1)} \times \frac{(3a + 1)}{(3a - 2)} = \frac{7a(2 - 3a)}{(3a + 1)} \times \frac{1}{(3a - 2)}$$ 5. **Multiply the remaining factors:** $$\frac{7a(2 - 3a)}{(3a + 1)(3a - 2)}$$ 6. **Rewrite $2 - 3a$ as $-(3a - 2)$ to simplify sign:** $$7a(2 - 3a) = 7a \times (-(3a - 2)) = -7a(3a - 2)$$ 7. **Substitute back:** $$\frac{-7a(3a - 2)}{(3a + 1)(3a - 2)}$$ 8. **Cancel $(3a - 2)$ numerator and denominator:** $$\frac{-7a\cancel{(3a - 2)}}{(3a + 1)\cancel{(3a - 2)}} = \frac{-7a}{3a + 1}$$ **Final answer:** $$\boxed{\frac{-7a}{3a + 1}}$$