Multiply $ \dfrac{6x-12}{x^2+7x+12} $ by $ \dfrac{2x^2+7x+3}{2x^2-4x} $ to get $ \dfrac{ 6x+3 }{ x^2+4x } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{6x-12}{x^2+7x+12} \cdot \frac{2x^2+7x+3}{2x^2-4x} & \xlongequal{\text{Step 1}} \frac{ 3 \cdot \color{blue}{ \left( 2x-4 \right) } }{ \left( x+4 \right) \cdot \color{red}{ \left( x+3 \right) } } \cdot \frac{ \left( 2x+1 \right) \cdot \color{red}{ \left( x+3 \right) } }{ x \cdot \color{blue}{ \left( 2x-4 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 3 }{ x+4 } \cdot \frac{ 2x+1 }{ x } \xlongequal{\text{Step 3}} \frac{ 3 \cdot \left( 2x+1 \right) }{ \left( x+4 \right) \cdot x } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 6x+3 }{ x^2+4x } \end{aligned} $$