Subtract $ \dfrac{7}{x} $ from $ 3x+2 $ to get $ \dfrac{ \color{purple}{ 3x^2+2x-7 } }{ x }$.

Step 1: Write $ 3x+2 $ 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 $\color{blue}{ x }$.

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

Subtract $4$ from $ \dfrac{3x^2+2x-7}{x} $ to get $ \dfrac{ \color{purple}{ 3x^2-2x-7 } }{ x }$.

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

$$ \begin{aligned} \frac{3x^2+2x-7}{x} -4 & \xlongequal{\text{Step 1}} \frac{3x^2+2x-7}{x} - \frac{4}{\color{red}{1}} = \frac{ 3x^2+2x-7 }{ x } - \frac{ 4 \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x^2+2x-7 } }{ x } - \frac{ \color{purple}{ 4x } }{ x }=\frac{ \color{purple}{ 3x^2-2x-7 } }{ x } \end{aligned} $$