Subtract $ \dfrac{1}{4} $ from $ 3x $ to get $ \dfrac{ \color{purple}{ 12x-1 } }{ 4 }$.

Step 1: Write $ 3x $ 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 $\color{blue}{ 4 }$.

$$ \begin{aligned} 3x- \frac{1}{4} & \xlongequal{\text{Step 1}} \frac{3x}{\color{red}{1}} - \frac{1}{4} = \frac{ 3x \cdot \color{blue}{ 4 }}{ 1 \cdot \color{blue}{ 4 }} - \frac{ 1 }{ 4 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 12x } }{ 4 } - \frac{ \color{purple}{ 1 } }{ 4 }=\frac{ \color{purple}{ 12x-1 } }{ 4 } \end{aligned} $$

Multiply $ \dfrac{12x-1}{4} $ by $ 4x+8 $ to get $ \dfrac{48x^2+92x-8}{4} $.

Step 1: Write $ 4x+8 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{12x-1}{4} \cdot 4x+8 & \xlongequal{\text{Step 1}} \frac{12x-1}{4} \cdot \frac{4x+8}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 12x-1 \right) \cdot \left( 4x+8 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 48x^2+96x-4x-8 }{ 4 } = \frac{48x^2+92x-8}{4} \end{aligned} $$