Question
Answer
$$(-3-6i)\cdot(1-9i)=54i^2+21i-3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-3-6i)\cdot(1-9i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-3+27i-6i+54i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}54i^2+21i-3\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{-3-6i}\right) $ by each term in $ \left( 1-9i\right) $. $$ \left( \color{blue}{-3-6i}\right) \cdot \left( 1-9i\right) = -3+27i-6i+54i^2 $$ |
| ② | Combine like terms: $$ -3+ \color{blue}{27i} \color{blue}{-6i} +54i^2 = 54i^2+ \color{blue}{21i} -3 $$ |
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