Tap the blue circles to see an explanation.
$$ \begin{aligned}(3-t)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9-6t+t^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}t^2-6t+9\end{aligned} $$ | |
① | Find $ \left(3-t\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3 } $ and $ B = \color{red}{ t }$. $$ \begin{aligned}\left(3-t\right)^2 = \color{blue}{3^2} -2 \cdot 3 \cdot t + \color{red}{t^2} = 9-6t+t^2\end{aligned} $$ |
② | Combine like terms: $$ t^2-6t+9 = t^2-6t+9 $$ |