Question
Answer
$$(x+4y)^2-(x+y)(x-y)=8xy+17y^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+4y)^2-(x+y)(x-y)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+8xy+16y^2-(x+y)(x-y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2+8xy+16y^2-(x^2-xy+xy-y^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^2+8xy+16y^2-(x^2-y^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^2+8xy+16y^2-x^2+y^2 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{x^2}+8xy+16y^2 -\cancel{x^2}+y^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}8xy+17y^2\end{aligned} $$ | |
| ① | Find $ \left(x+4y\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}{ 4y }$. $$ \begin{aligned}\left(x+4y\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 4y + \color{red}{\left( 4y \right)^2} = x^2+8xy+16y^2\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x+y}\right) $ by each term in $ \left( x-y\right) $. $$ \left( \color{blue}{x+y}\right) \cdot \left( x-y\right) = x^2 -\cancel{xy}+ \cancel{xy}-y^2 $$ |
| ③ | Combine like terms: $$ x^2 \, \color{blue}{ -\cancel{xy}} \,+ \, \color{blue}{ \cancel{xy}} \,-y^2 = x^2-y^2 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^2-y^2 \right) = -x^2+y^2 $$ |
| ⑤ | Combine like terms: $$ \, \color{blue}{ \cancel{x^2}} \,+8xy+ \color{green}{16y^2} \, \color{blue}{ -\cancel{x^2}} \,+ \color{green}{y^2} = 8xy+ \color{green}{17y^2} $$ |
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