| $$ \begin{aligned}\frac{4x^2-25}{2x^2+x-15}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x+5}{x+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^2-25}{2x^2+x-15} $ to $ \dfrac{2x+5}{x+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-5}$. $$ \begin{aligned} \frac{4x^2-25}{2x^2+x-15} & =\frac{ \left( 2x+5 \right) \cdot \color{blue}{ \left( 2x-5 \right) }}{ \left( x+3 \right) \cdot \color{blue}{ \left( 2x-5 \right) }} = \\[1ex] &= \frac{2x+5}{x+3} \end{aligned} $$ |