Tap the blue circles to see an explanation.
| $$ \begin{aligned}2x^2-(2x+7)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2x^2-(4x^2+28x+49) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x^2-4x^2-28x-49 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-2x^2-28x-49\end{aligned} $$ | |
| ① | Find $ \left(2x+7\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}{ 7 }$. $$ \begin{aligned}\left(2x+7\right)^2 = \color{blue}{\left( 2x \right)^2} +2 \cdot 2x \cdot 7 + \color{red}{7^2} = 4x^2+28x+49\end{aligned} $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4x^2+28x+49 \right) = -4x^2-28x-49 $$ |
| ③ | Combine like terms: $$ \color{blue}{2x^2} \color{blue}{-4x^2} -28x-49 = \color{blue}{-2x^2} -28x-49 $$ |