Simplify $ \dfrac{x^2+5x+4}{x^2+2x+1} $ to $ \dfrac{x+4}{x+1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+1}$.

$$ \begin{aligned} \frac{x^2+5x+4}{x^2+2x+1} & =\frac{ \left( x+4 \right) \cdot \color{blue}{ \left( x+1 \right) }}{ \left( x+1 \right) \cdot \color{blue}{ \left( x+1 \right) }} = \\[1ex] &= \frac{x+4}{x+1} \end{aligned} $$

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

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

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x+4}{x+1} \cdot 2x+2 & \xlongequal{\text{Step 1}} \frac{x+4}{x+1} \cdot \frac{2x+2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ x+4 }{ 1 \cdot \color{red}{ \left( x+1 \right) } } \cdot \frac{ 2 \cdot \color{red}{ \left( x+1 \right) } }{ 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x+4 }{ 1 } \cdot \frac{ 2 }{ 1 } \xlongequal{\text{Step 4}} \frac{ \left( x+4 \right) \cdot 2 }{ 1 \cdot 1 } \xlongequal{\text{Step 5}} \frac{ 2x+8 }{ 1 } = \\[1ex] &=2x+8 \end{aligned} $$