Answer
$$4n(3n^2+n-4)-n\cdot(3-n)=12n^3+5n^2-19n$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}4n(3n^2+n-4)-n\cdot(3-n)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12n^3+4n^2-16n-(3n-n^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12n^3+4n^2-16n-3n+n^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}12n^3+5n^2-19n\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{4n} $ by $ \left( 3n^2+n-4\right) $ $$ \color{blue}{4n} \cdot \left( 3n^2+n-4\right) = 12n^3+4n^2-16n $$Multiply $ \color{blue}{n} $ by $ \left( 3-n\right) $ $$ \color{blue}{n} \cdot \left( 3-n\right) = 3n-n^2 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3n-n^2 \right) = -3n+n^2 $$ |
| ③ | Combine like terms: $$ 12n^3+ \color{blue}{4n^2} \color{red}{-16n} \color{red}{-3n} + \color{blue}{n^2} = 12n^3+ \color{blue}{5n^2} \color{red}{-19n} $$ |
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