Multiply each term of $ \left( \color{blue}{x-2}\right) $ by each term in $ \left( x+4\right) $.

$$ \left( \color{blue}{x-2}\right) \cdot \left( x+4\right) = x^2+4x-2x-8 $$

Combine like terms:

$$ x^2+ \color{blue}{4x} \color{blue}{-2x} -8 = x^2+ \color{blue}{2x} -8 $$

Simplify $ \dfrac{x-2}{x^2+2x-8} $ to $ \dfrac{1}{x+4} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-2}$.

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

Multiply $5$ by $ \dfrac{1}{x+4} $ to get $ \dfrac{ 5 }{ x+4 } $.

Step 1: Write $ 5 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 5 \cdot \frac{1}{x+4} & \xlongequal{\text{Step 1}} \frac{5}{\color{red}{1}} \cdot \frac{1}{x+4} \xlongequal{\text{Step 2}} \frac{ 5 \cdot 1 }{ 1 \cdot \left( x+4 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5 }{ x+4 } \end{aligned} $$