Multiply each term of $ \left( \color{blue}{x-4i}\right) $ by each term in $ \left( x+4i\right) $.

$$ \left( \color{blue}{x-4i}\right) \cdot \left( x+4i\right) = x^2+ \cancel{4ix} -\cancel{4ix}-16i^2 $$

Combine like terms:

$$ x^2+ \, \color{blue}{ \cancel{4ix}} \, \, \color{blue}{ -\cancel{4ix}} \,-16i^2 = -16i^2+x^2 $$

$$ -16i^2 = -16 \cdot (-1) = 16 $$

Subtract $ \dfrac{3}{5} $ from $ x $ to get $ \dfrac{ \color{purple}{ 5x-3 } }{ 5 }$.

Step 1: Write $ x $ 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 first fraction by $\color{blue}{ 5 }$.

$$ \begin{aligned} x- \frac{3}{5} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} - \frac{3}{5} = \frac{ x \cdot \color{blue}{ 5 }}{ 1 \cdot \color{blue}{ 5 }} - \frac{ 3 }{ 5 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 5x } }{ 5 } - \frac{ \color{purple}{ 3 } }{ 5 }=\frac{ \color{purple}{ 5x-3 } }{ 5 } \end{aligned} $$

Multiply $16+x^2$ by $ \dfrac{5x-3}{5} $ to get $ \dfrac{5x^3-3x^2+80x-48}{5} $.

Step 1: Write $ 16+x^2 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 16+x^2 \cdot \frac{5x-3}{5} & \xlongequal{\text{Step 1}} \frac{16+x^2}{\color{red}{1}} \cdot \frac{5x-3}{5} \xlongequal{\text{Step 2}} \frac{ \left( 16+x^2 \right) \cdot \left( 5x-3 \right) }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 80x-48+5x^3-3x^2 }{ 5 } = \frac{5x^3-3x^2+80x-48}{5} \end{aligned} $$