Multiply $ \dfrac{x^2+3x-4}{x^2+4x+4} $ by $ \dfrac{2x^2+4x}{x^2-4x+3} $ to get $ \dfrac{2x^2+8x}{x^2-x-6} $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

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