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

Step 1: Write $ x+4 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Multiply $ \dfrac{2x+8}{3} $ by $ x+1 $ to get $ \dfrac{2x^2+10x+8}{3} $.

Step 1: Write $ x+1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2x+8}{3} \cdot x+1 & \xlongequal{\text{Step 1}} \frac{2x+8}{3} \cdot \frac{x+1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 2x+8 \right) \cdot \left( x+1 \right) }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^2+2x+8x+8 }{ 3 } = \frac{2x^2+10x+8}{3} \end{aligned} $$

Multiply $ \dfrac{2x^2+10x+8}{3} $ by $ x-3 $ to get $ \dfrac{2x^3+4x^2-22x-24}{3} $.

Step 1: Write $ x-3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2x^2+10x+8}{3} \cdot x-3 & \xlongequal{\text{Step 1}} \frac{2x^2+10x+8}{3} \cdot \frac{x-3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 2x^2+10x+8 \right) \cdot \left( x-3 \right) }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^3-6x^2+10x^2-30x+8x-24 }{ 3 } = \frac{2x^3+4x^2-22x-24}{3} \end{aligned} $$