Simplify $ \dfrac{5x^2-20}{5x^2+10x} $ to $ \dfrac{x-2}{x} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{5x+10}$.

$$ \begin{aligned} \frac{5x^2-20}{5x^2+10x} & =\frac{ \left( x-2 \right) \cdot \color{blue}{ \left( 5x+10 \right) }}{ x \cdot \color{blue}{ \left( 5x+10 \right) }} = \\[1ex] &= \frac{x-2}{x} \end{aligned} $$

Simplify $ \dfrac{x^2+2x-8}{2x^2+9x+4} $ to $ \dfrac{x-2}{2x+1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+4}$.

$$ \begin{aligned} \frac{x^2+2x-8}{2x^2+9x+4} & =\frac{ \left( x-2 \right) \cdot \color{blue}{ \left( x+4 \right) }}{ \left( 2x+1 \right) \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &= \frac{x-2}{2x+1} \end{aligned} $$

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

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

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