Multiply $ \dfrac{3c-4}{2c^2+5c+3} $ by $ 6c^4+15c^3+9c^2 $ to get $ 9c^3-12c^2$.

Step 1: Write $ 6c^4+15c^3+9c^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{3c-4}{2c^2+5c+3} \cdot 6c^4+15c^3+9c^2 & \xlongequal{\text{Step 1}} \frac{3c-4}{2c^2+5c+3} \cdot \frac{6c^4+15c^3+9c^2}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 3c-4 }{ 1 \cdot \color{red}{ \left( 2c^2+5c+3 \right) } } \cdot \frac{ 3c^2 \cdot \color{red}{ \left( 2c^2+5c+3 \right) } }{ 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3c-4 }{ 1 } \cdot \frac{ 3c^2 }{ 1 } \xlongequal{\text{Step 4}} \frac{ \left( 3c-4 \right) \cdot 3c^2 }{ 1 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 5}} \frac{ 9c^3-12c^2 }{ 1 } =9c^3-12c^2 \end{aligned} $$