Answer
$$(5n-5)\cdot(2+2n)=10n^2-10$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(5n-5)\cdot(2+2n)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}10n+10n^2-10-10n \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{10n}+10n^2-10 -\cancel{10n} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}10n^2-10\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{5n-5}\right) $ by each term in $ \left( 2+2n\right) $. $$ \left( \color{blue}{5n-5}\right) \cdot \left( 2+2n\right) = \cancel{10n}+10n^2-10 -\cancel{10n} $$ |
| ② | Combine like terms: $$ \, \color{blue}{ \cancel{10n}} \,+10n^2-10 \, \color{blue}{ -\cancel{10n}} \, = 10n^2-10 $$ |
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