Answer
$$(n-9)(n^2+4n-10)=n^3-5n^2-46n+90$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n-9)(n^2+4n-10)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}n^3+4n^2-10n-9n^2-36n+90 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}n^3-5n^2-46n+90\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{n-9}\right) $ by each term in $ \left( n^2+4n-10\right) $. $$ \left( \color{blue}{n-9}\right) \cdot \left( n^2+4n-10\right) = n^3+4n^2-10n-9n^2-36n+90 $$ |
| ② | Combine like terms: $$ n^3+ \color{blue}{4n^2} \color{red}{-10n} \color{blue}{-9n^2} \color{red}{-36n} +90 = n^3 \color{blue}{-5n^2} \color{red}{-46n} +90 $$ |
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