Multiply $ \dfrac{9a^2+9a+2}{4a^2+8a+3} $ by $ \dfrac{2a^2+15a+7}{9a^2-4} $ to get $ \dfrac{3a^2+22a+7}{6a^2+5a-6} $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{9a^2+9a+2}{4a^2+8a+3} \cdot \frac{2a^2+15a+7}{9a^2-4} & \xlongequal{\text{Step 1}} \frac{ \left( 3a+1 \right) \cdot \color{blue}{ \left( 3a+2 \right) } }{ \left( 2a+3 \right) \cdot \color{red}{ \left( 2a+1 \right) } } \cdot \frac{ \left( a+7 \right) \cdot \color{red}{ \left( 2a+1 \right) } }{ \left( 3a-2 \right) \cdot \color{blue}{ \left( 3a+2 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 3a+1 }{ 2a+3 } \cdot \frac{ a+7 }{ 3a-2 } \xlongequal{\text{Step 3}} \frac{ \left( 3a+1 \right) \cdot \left( a+7 \right) }{ \left( 2a+3 \right) \cdot \left( 3a-2 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 3a^2+21a+a+7 }{ 6a^2-4a+9a-6 } = \frac{3a^2+22a+7}{6a^2+5a-6} \end{aligned} $$