Answer
$$(5-4r)(6+3r-2r^2)=8r^3-22r^2-9r+30$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(5-4r)(6+3r-2r^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}30+15r-10r^2-24r-12r^2+8r^3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}8r^3-22r^2-9r+30\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{5-4r}\right) $ by each term in $ \left( 6+3r-2r^2\right) $. $$ \left( \color{blue}{5-4r}\right) \cdot \left( 6+3r-2r^2\right) = 30+15r-10r^2-24r-12r^2+8r^3 $$ |
| ② | Combine like terms: $$ 30+ \color{blue}{15r} \color{red}{-10r^2} \color{blue}{-24r} \color{red}{-12r^2} +8r^3 = 8r^3 \color{red}{-22r^2} \color{blue}{-9r} +30 $$ |
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