Answer
$$(4m+n)(2m^2-3mn+8n^2)=8m^3-10m^2n+29mn^2+8n^3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4m+n)(2m^2-3mn+8n^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}8m^3-12m^2n+32mn^2+2m^2n-3mn^2+8n^3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}8m^3-10m^2n+29mn^2+8n^3\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4m+n}\right) $ by each term in $ \left( 2m^2-3mn+8n^2\right) $. $$ \left( \color{blue}{4m+n}\right) \cdot \left( 2m^2-3mn+8n^2\right) = 8m^3-12m^2n+32mn^2+2m^2n-3mn^2+8n^3 $$ |
| ② | Combine like terms: $$ 8m^3 \color{blue}{-12m^2n} + \color{red}{32mn^2} + \color{blue}{2m^2n} \color{red}{-3mn^2} +8n^3 = 8m^3 \color{blue}{-10m^2n} + \color{red}{29mn^2} +8n^3 $$ |
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