| $$ \begin{aligned}\frac{x^2+12x+35}{x+5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x+7\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+12x+35}{x+5} $ to $ x+7$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+5}$. $$ \begin{aligned} \frac{x^2+12x+35}{x+5} & =\frac{ \left( x+7 \right) \cdot \color{blue}{ \left( x+5 \right) }}{ 1 \cdot \color{blue}{ \left( x+5 \right) }} = \\[1ex] &= \frac{x+7}{1} =x+7 \end{aligned} $$ |