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