Multiply $ \dfrac{3}{x} $ by $ x+2 $ to get $ \dfrac{ 3x+6 }{ x } $.

Step 1: Write $ 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.

$$ \begin{aligned} \frac{3}{x} \cdot x+2 & \xlongequal{\text{Step 1}} \frac{3}{x} \cdot \frac{x+2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 3 \cdot \left( x+2 \right) }{ x \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x+6 }{ x } \end{aligned} $$

Subtract $ \dfrac{3x+6}{x} $ from $ x^2+2x $ to get $ \dfrac{ \color{purple}{ x^3+2x^2-3x-6 } }{ x }$.

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

$$ \begin{aligned} x^2+2x- \frac{3x+6}{x} & \xlongequal{\text{Step 1}} \frac{x^2+2x}{\color{red}{1}} - \frac{3x+6}{x} = \frac{ \left( x^2+2x \right) \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} - \frac{ 3x+6 }{ x } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^3+2x^2 } }{ x } - \frac{ \color{purple}{ 3x+6 } }{ x }=\frac{ \color{purple}{ x^3+2x^2 - \left( 3x+6 \right) } }{ x } = \\[1ex] &=\frac{ \color{purple}{ x^3+2x^2-3x-6 } }{ x } \end{aligned} $$