| $$ \begin{aligned}\frac{x^3+7x^2+10x-6}{x+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+4x-2\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^3+7x^2+10x-6}{x+3} $ to $ x^2+4x-2$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+3}$. $$ \begin{aligned} \frac{x^3+7x^2+10x-6}{x+3} & =\frac{ \left( x^2+4x-2 \right) \cdot \color{blue}{ \left( x+3 \right) }}{ 1 \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &= \frac{x^2+4x-2}{1} =x^2+4x-2 \end{aligned} $$ |