Multiply $ \dfrac{2x^2-32}{4} $ by $ x $ to get $ \dfrac{ 2x^3-32x }{ 4 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2x^2-32}{4} \cdot x & \xlongequal{\text{Step 1}} \frac{2x^2-32}{4} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 2x^2-32 \right) \cdot x }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^3-32x }{ 4 } \end{aligned} $$

Subtract $16$ from $ \dfrac{2x^3-32x}{4} $ to get $ \dfrac{ \color{purple}{ 2x^3-32x-64 } }{ 4 }$.

Step 1: Write $ 16 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{4}$.

$$ \begin{aligned} \frac{2x^3-32x}{4} -16 & \xlongequal{\text{Step 1}} \frac{2x^3-32x}{4} - \frac{16}{\color{red}{1}} = \frac{ 2x^3-32x }{ 4 } - \frac{ 16 \cdot \color{blue}{ 4 }}{ 1 \cdot \color{blue}{ 4 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^3-32x } }{ 4 } - \frac{ \color{purple}{ 64 } }{ 4 }=\frac{ \color{purple}{ 2x^3-32x-64 } }{ 4 } \end{aligned} $$