Question
Answer
$$(3-2i)\cdot(5+i)=-2i^2-7i+15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2i)\cdot(5+i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}15+3i-10i-2i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-2i^2-7i+15\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-2i}\right) $ by each term in $ \left( 5+i\right) $. $$ \left( \color{blue}{3-2i}\right) \cdot \left( 5+i\right) = 15+3i-10i-2i^2 $$ |
| ② | Combine like terms: $$ 15+ \color{blue}{3i} \color{blue}{-10i} -2i^2 = -2i^2 \color{blue}{-7i} +15 $$ |
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