Simplify $ \dfrac{x^3+8x^2}{x^2+x-56} $ to $ \dfrac{x^2}{x-7} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+8}$.

$$ \begin{aligned} \frac{x^3+8x^2}{x^2+x-56} & =\frac{ x^2 \cdot \color{blue}{ \left( x+8 \right) }}{ \left( x-7 \right) \cdot \color{blue}{ \left( x+8 \right) }} = \\[1ex] &= \frac{x^2}{x-7} \end{aligned} $$

Multiply $ \dfrac{x^2}{x-7} $ by $ \dfrac{x^2-11x+28}{4x} $ to get $ \dfrac{ x^3-4x^2 }{ 4x } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^2}{x-7} \cdot \frac{x^2-11x+28}{4x} & \xlongequal{\text{Step 1}} \frac{ x^2 }{ 1 \cdot \color{red}{ \left( x-7 \right) } } \cdot \frac{ \left( x-4 \right) \cdot \color{red}{ \left( x-7 \right) } }{ 4x } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x^2 }{ 1 } \cdot \frac{ x-4 }{ 4x } \xlongequal{\text{Step 3}} \frac{ x^2 \cdot \left( x-4 \right) }{ 1 \cdot 4x } \xlongequal{\text{Step 4}} \frac{ x^3-4x^2 }{ 4x } \end{aligned} $$