Answer
$$(2x-8)(2x+8)=4x^2-64$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x-8)(2x+8)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2+16x-16x-64 \xlongequal{ } \\[1 em] & \xlongequal{ }4x^2+ \cancel{16x} -\cancel{16x}-64 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^2-64\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x-8}\right) $ by each term in $ \left( 2x+8\right) $. $$ \left( \color{blue}{2x-8}\right) \cdot \left( 2x+8\right) = 4x^2+ \cancel{16x} -\cancel{16x}-64 $$ |
| ② | Combine like terms: $$ 4x^2+ \, \color{blue}{ \cancel{16x}} \, \, \color{blue}{ -\cancel{16x}} \,-64 = 4x^2-64 $$ |
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