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