Multiply $ \dfrac{8x^3+4x^2}{x^2-5x+6} $ by $ \dfrac{x^2-9}{16x^2+8} $ to get $ \dfrac{2x^4+7x^3+3x^2}{4x^3-8x^2+2x-4} $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{8x^3+4x^2}{x^2-5x+6} \cdot \frac{x^2-9}{16x^2+8} & \xlongequal{\text{Step 1}} \frac{ \left( 2x^3+x^2 \right) \cdot \color{blue}{4} }{ \left( x-2 \right) \cdot \color{red}{ \left( x-3 \right) } } \cdot \frac{ \left( x+3 \right) \cdot \color{red}{ \left( x-3 \right) } }{ \left( 4x^2+2 \right) \cdot \color{blue}{4} } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 2x^3+x^2 }{ x-2 } \cdot \frac{ x+3 }{ 4x^2+2 } \xlongequal{\text{Step 3}} \frac{ \left( 2x^3+x^2 \right) \cdot \left( x+3 \right) }{ \left( x-2 \right) \cdot \left( 4x^2+2 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 2x^4+6x^3+x^3+3x^2 }{ 4x^3+2x-8x^2-4 } = \frac{2x^4+7x^3+3x^2}{4x^3-8x^2+2x-4} \end{aligned} $$