$$ \begin{aligned}(2x-3)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2-12x+9\end{aligned} $$ | |
① | Find $ \left(2x-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(2x-3\right)^2 = \color{blue}{\left( 2x \right)^2} -2 \cdot 2x \cdot 3 + \color{red}{3^2} = 4x^2-12x+9\end{aligned} $$ |