Question
Answer
$$(2+3i)\cdot(6+7i)=21i^2+32i+12$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2+3i)\cdot(6+7i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12+14i+18i+21i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}21i^2+32i+12\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2+3i}\right) $ by each term in $ \left( 6+7i\right) $. $$ \left( \color{blue}{2+3i}\right) \cdot \left( 6+7i\right) = 12+14i+18i+21i^2 $$ |
| ② | Combine like terms: $$ 12+ \color{blue}{14i} + \color{blue}{18i} +21i^2 = 21i^2+ \color{blue}{32i} +12 $$ |
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