Tap the blue circles to see an explanation.
$$ \begin{aligned}2(x+y)^2-(x+y)(y-2x)-(y-2x)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2(x^2+2xy+y^2)-(x+y)(y-2x)-(1y^2-4xy+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x^2+4xy+2y^2-(xy-2x^2+y^2-2xy)-(1y^2-4xy+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2x^2+4xy+2y^2-(-2x^2-xy+y^2)-(1y^2-4xy+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2x^2+4xy+2y^2+2x^2+xy-y^2-(1y^2-4xy+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4x^2+5xy+y^2-(1y^2-4xy+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}4x^2+5xy+y^2-y^2+4xy-4x^2 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{4x^2}+5xy+ \cancel{y^2} -\cancel{y^2}+4xy -\cancel{4x^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}9xy\end{aligned} $$ | |
① | Find $ \left(x+y\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}{ y }$. $$ \begin{aligned}\left(x+y\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot y + \color{red}{y^2} = x^2+2xy+y^2\end{aligned} $$Find $ \left(y-2x\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ y } $ and $ B = \color{red}{ 2x }$. $$ \begin{aligned}\left(y-2x\right)^2 = \color{blue}{y^2} -2 \cdot y \cdot 2x + \color{red}{\left( 2x \right)^2} = y^2-4xy+4x^2\end{aligned} $$ |
② | Multiply $ \color{blue}{2} $ by $ \left( x^2+2xy+y^2\right) $ $$ \color{blue}{2} \cdot \left( x^2+2xy+y^2\right) = 2x^2+4xy+2y^2 $$ Multiply each term of $ \left( \color{blue}{x+y}\right) $ by each term in $ \left( y-2x\right) $. $$ \left( \color{blue}{x+y}\right) \cdot \left( y-2x\right) = xy-2x^2+y^2-2xy $$ |
③ | Combine like terms: $$ \color{blue}{xy} -2x^2+y^2 \color{blue}{-2xy} = -2x^2 \color{blue}{-xy} +y^2 $$ |
④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( -2x^2-xy+y^2 \right) = 2x^2+xy-y^2 $$ |
⑤ | Combine like terms: $$ \color{blue}{2x^2} + \color{red}{4xy} + \color{green}{2y^2} + \color{blue}{2x^2} + \color{red}{xy} \color{green}{-y^2} = \color{blue}{4x^2} + \color{red}{5xy} + \color{green}{y^2} $$ |
⑥ | Remove the parentheses by changing the sign of each term within them. $$ - \left( y^2-4xy+4x^2 \right) = -y^2+4xy-4x^2 $$ |
⑦ | Combine like terms: $$ \, \color{blue}{ \cancel{4x^2}} \,+ \color{green}{5xy} + \, \color{orange}{ \cancel{y^2}} \, \, \color{orange}{ -\cancel{y^2}} \,+ \color{green}{4xy} \, \color{blue}{ -\cancel{4x^2}} \, = \color{green}{9xy} $$ |