Question
Answer
$$(3r-8)(11r-4n)=-12nr+33r^2+32n-88r$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3r-8)(11r-4n)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}33r^2-12nr-88r+32n \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-12nr+33r^2+32n-88r\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3r-8}\right) $ by each term in $ \left( 11r-4n\right) $. $$ \left( \color{blue}{3r-8}\right) \cdot \left( 11r-4n\right) = 33r^2-12nr-88r+32n $$ |
| ② | Combine like terms: $$ -12nr+33r^2+32n-88r = -12nr+33r^2+32n-88r $$ |
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