Simplify $ \dfrac{x^2-25}{x-5} $ to $ x+5$.

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

$$ \begin{aligned} \frac{x^2-25}{x-5} & =\frac{ \left( x+5 \right) \cdot \color{blue}{ \left( x-5 \right) }}{ 1 \cdot \color{blue}{ \left( x-5 \right) }} = \\[1ex] &= \frac{x+5}{1} =x+5 \end{aligned} $$

Multiply each term of $ \left( \color{blue}{x+5}\right) $ by each term in $ \left( x+3\right) $.

$$ \left( \color{blue}{x+5}\right) \cdot \left( x+3\right) = x^2+3x+5x+15 $$

Combine like terms:

$$ x^2+ \color{blue}{3x} + \color{blue}{5x} +15 = x^2+ \color{blue}{8x} +15 $$

Multiply $x^2+8x+15$ by $ \dfrac{x^2+7x+1}{x+4} $ to get $ \dfrac{x^4+15x^3+72x^2+113x+15}{x+4} $.

Step 1: Write $ x^2+8x+15 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} x^2+8x+15 \cdot \frac{x^2+7x+1}{x+4} & \xlongequal{\text{Step 1}} \frac{x^2+8x+15}{\color{red}{1}} \cdot \frac{x^2+7x+1}{x+4} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^2+8x+15 \right) \cdot \left( x^2+7x+1 \right) }{ 1 \cdot \left( x+4 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4+7x^3+x^2+8x^3+56x^2+8x+15x^2+105x+15 }{ x+4 } = \frac{x^4+15x^3+72x^2+113x+15}{x+4} \end{aligned} $$