$$ \begin{aligned}(4x+15)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}16x^2+120x+225\end{aligned} $$ | |
① | Find $ \left(4x+15\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 4x } $ and $ B = \color{red}{ 15 }$. $$ \begin{aligned}\left(4x+15\right)^2 = \color{blue}{\left( 4x \right)^2} +2 \cdot 4x \cdot 15 + \color{red}{15^2} = 16x^2+120x+225\end{aligned} $$ |