Add $ \dfrac{3x^3-4x^2+1}{x} $ and $ 2 $ to get $ \dfrac{ \color{purple}{ 3x^3-4x^2+2x+1 } }{ x }$.

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

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

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