Tap the blue circles to see an explanation.
$$ \begin{aligned}2(x-3)^2-8& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2(x^2-6x+9)-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x^2-12x+18-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2x^2-12x+10\end{aligned} $$ | |
① | Find $ \left(x-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}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x-3\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 3 + \color{red}{3^2} = x^2-6x+9\end{aligned} $$ |
② | Multiply $ \color{blue}{2} $ by $ \left( x^2-6x+9\right) $ $$ \color{blue}{2} \cdot \left( x^2-6x+9\right) = 2x^2-12x+18 $$ |
③ | Combine like terms: $$ 2x^2-12x+ \color{blue}{18} \color{blue}{-8} = 2x^2-12x+ \color{blue}{10} $$ |