Answer
$$(n-2m)(n^2+4m)=-2mn^2+n^3-8m^2+4mn$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n-2m)(n^2+4m)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}n^3+4mn-2mn^2-8m^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-2mn^2+n^3-8m^2+4mn\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{n-2m}\right) $ by each term in $ \left( n^2+4m\right) $. $$ \left( \color{blue}{n-2m}\right) \cdot \left( n^2+4m\right) = n^3+4mn-2mn^2-8m^2 $$ |
| ② | Combine like terms: $$ -2mn^2+n^3-8m^2+4mn = -2mn^2+n^3-8m^2+4mn $$ |
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