Divide both the top and bottom numbers by $ \color{orangered}{ 4 } $.

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

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.

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

Add $3x^3-15x$ and $ \dfrac{2x^2}{3} $ to get $ \dfrac{ \color{purple}{ 9x^3+2x^2-45x } }{ 3 }$.

Step 1: Write $ 3x^3-15x $ 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 first fraction by $\color{blue}{ 3 }$.

$$ \begin{aligned} 3x^3-15x+ \frac{2x^2}{3} & \xlongequal{\text{Step 1}} \frac{3x^3-15x}{\color{red}{1}} + \frac{2x^2}{3} = \frac{ \left( 3x^3-15x \right) \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} + \frac{ 2x^2 }{ 3 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 9x^3-45x } }{ 3 } + \frac{ \color{purple}{ 2x^2 } }{ 3 }=\frac{ \color{purple}{ 9x^3+2x^2-45x } }{ 3 } \end{aligned} $$