Find $ \left(x+2\right)^2 $ using formula.

$$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 2 }$.

$$ \begin{aligned}\left(x+2\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 2 + \color{red}{2^2} = x^2+4x+4\end{aligned} $$

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

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

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