Multiply $ \dfrac{3}{8} $ by $ x^2 $ to get $ \dfrac{ 3x^2 }{ 8 } $.

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.

$$ \begin{aligned} \frac{3}{8} \cdot x^2 & \xlongequal{\text{Step 1}} \frac{3}{8} \cdot \frac{x^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 3 \cdot x^2 }{ 8 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x^2 }{ 8 } \end{aligned} $$

Multiply $ \dfrac{1}{4} $ by $ x $ to get $ \dfrac{ x }{ 4 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{4} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{4} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 4 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 4 } \end{aligned} $$

Multiply $ \dfrac{1}{3} $ by $ x $ to get $ \dfrac{ x }{ 3 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 3 } \end{aligned} $$

Add $ \dfrac{3x^2}{8} $ and $ \dfrac{x}{4} $ to get $ \dfrac{ \color{purple}{ 3x^2+2x } }{ 8 }$.

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

$$ \begin{aligned} \frac{3x^2}{8} + \frac{x}{4} & = \frac{ 3x^2 }{ 8 } + \frac{ x \cdot \color{blue}{ 2 }}{ 4 \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ 3x^2 } }{ 8 } + \frac{ \color{purple}{ 2x } }{ 8 }=\frac{ \color{purple}{ 3x^2+2x } }{ 8 } \end{aligned} $$

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

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

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

Subtract $ \dfrac{2}{5} $ from $ \dfrac{3x^2+2x}{8} $ to get $ \dfrac{ \color{purple}{ 15x^2+10x-16 } }{ 40 }$.

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}{ 5 }$ and the second by $\color{blue}{ 8 }$.

$$ \begin{aligned} \frac{3x^2+2x}{8} - \frac{2}{5} & = \frac{ \left( 3x^2+2x \right) \cdot \color{blue}{ 5 }}{ 8 \cdot \color{blue}{ 5 }} - \frac{ 2 \cdot \color{blue}{ 8 }}{ 5 \cdot \color{blue}{ 8 }} = \\[1ex] &=\frac{ \color{purple}{ 15x^2+10x } }{ 40 } - \frac{ \color{purple}{ 16 } }{ 40 }=\frac{ \color{purple}{ 15x^2+10x-16 } }{ 40 } \end{aligned} $$

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

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 $\color{blue}{3}$.

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

Multiply $ \dfrac{15x^2+10x-16}{40} $ by $ \dfrac{6x^3-x+6}{3} $ to get $ \dfrac{90x^5+60x^4-111x^3+80x^2+76x-96}{120} $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{15x^2+10x-16}{40} \cdot \frac{6x^3-x+6}{3} & \xlongequal{\text{Step 1}} \frac{ \left( 15x^2+10x-16 \right) \cdot \left( 6x^3-x+6 \right) }{ 40 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 90x^5-15x^3+90x^2+60x^4-10x^2+60x-96x^3+16x-96 }{ 120 } = \frac{90x^5+60x^4-111x^3+80x^2+76x-96}{120} \end{aligned} $$