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