| $$ \begin{aligned}x\cdot4(x-15)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2-60x\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{4x} $ by $ \left( x-15\right) $ $$ \color{blue}{4x} \cdot \left( x-15\right) = 4x^2-60x $$ |