| $$ \begin{aligned}\frac{2x^2-32}{4x-16}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x+4}{2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^2-32}{4x-16} $ to $ \dfrac{x+4}{2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-8}$. $$ \begin{aligned} \frac{2x^2-32}{4x-16} & =\frac{ \left( x+4 \right) \cdot \color{blue}{ \left( 2x-8 \right) }}{ 2 \cdot \color{blue}{ \left( 2x-8 \right) }} = \\[1ex] &= \frac{x+4}{2} \end{aligned} $$ |