Multiply each term of $ \left( \color{blue}{x-2}\right) $ by each term in $ \left( x+2\right) $.

$$ \left( \color{blue}{x-2}\right) \cdot \left( x+2\right) = x^2+ \cancel{2x} -\cancel{2x}-4 $$

Combine like terms:

$$ x^2+ \, \color{blue}{ \cancel{2x}} \, \, \color{blue}{ -\cancel{2x}} \,-4 = x^2-4 $$

Subtract $ \dfrac{1}{2} $ from $ x $ to get $ \dfrac{ \color{purple}{ 2x-1 } }{ 2 }$.

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

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

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

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

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

Multiply $x^2-4$ by $ \dfrac{2x-1}{2} $ to get $ \dfrac{ 2x^3-x^2-8x+4 }{ 2 } $.

Step 1: Write $ x^2-4 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} x^2-4 \cdot \frac{2x-1}{2} & \xlongequal{\text{Step 1}} \frac{x^2-4}{\color{red}{1}} \cdot \frac{2x-1}{2} \xlongequal{\text{Step 2}} \frac{ \left( x^2-4 \right) \cdot \left( 2x-1 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^3-x^2-8x+4 }{ 2 } \end{aligned} $$

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

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

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

Multiply $ \dfrac{2x^3-x^2-8x+4}{2} $ by $ \dfrac{2x+1}{2} $ to get $ \dfrac{4x^4-17x^2+4}{4} $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2x^3-x^2-8x+4}{2} \cdot \frac{2x+1}{2} & \xlongequal{\text{Step 1}} \frac{ \left( 2x^3-x^2-8x+4 \right) \cdot \left( 2x+1 \right) }{ 2 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 4x^4+ \cancel{2x^3} -\cancel{2x^3}-x^2-16x^2 -\cancel{8x}+ \cancel{8x}+4 }{ 4 } = \frac{4x^4-17x^2+4}{4} \end{aligned} $$