Simplify $ \dfrac{2x^5+2x^4-4x^3}{6x^3-48x^2+42x} $ to $ \dfrac{x^3+2x^2}{3x-21} $.

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

$$ \begin{aligned} \frac{2x^5+2x^4-4x^3}{6x^3-48x^2+42x} & =\frac{ \left( x^3+2x^2 \right) \cdot \color{blue}{ \left( 2x^2-2x \right) }}{ \left( 3x-21 \right) \cdot \color{blue}{ \left( 2x^2-2x \right) }} = \\[1ex] &= \frac{x^3+2x^2}{3x-21} \end{aligned} $$

Find $ \left(x-7\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 7 }$.

$$ \begin{aligned}\left(x-7\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 7 + \color{red}{7^2} = x^2-14x+49\end{aligned} $$

Multiply $ \dfrac{x^3+2x^2}{3x-21} $ by $ \dfrac{x^2-14x+49}{x+2} $ to get $ \dfrac{ x^3-7x^2 }{ 3 } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^3+2x^2}{3x-21} \cdot \frac{x^2-14x+49}{x+2} & \xlongequal{\text{Step 1}} \frac{ x^2 \cdot \color{blue}{ \left( x+2 \right) } }{ 3 \cdot \color{red}{ \left( x-7 \right) } } \cdot \frac{ \left( x-7 \right) \cdot \color{red}{ \left( x-7 \right) } }{ 1 \cdot \color{blue}{ \left( x+2 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x^2 }{ 3 } \cdot \frac{ x-7 }{ 1 } \xlongequal{\text{Step 3}} \frac{ x^2 \cdot \left( x-7 \right) }{ 3 \cdot 1 } \xlongequal{\text{Step 4}} \frac{ x^3-7x^2 }{ 3 } \end{aligned} $$