Question
Answer
$$(3-4i)\cdot(5+2i)=-8i^2-14i+15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-4i)\cdot(5+2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}15+6i-20i-8i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-8i^2-14i+15\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-4i}\right) $ by each term in $ \left( 5+2i\right) $. $$ \left( \color{blue}{3-4i}\right) \cdot \left( 5+2i\right) = 15+6i-20i-8i^2 $$ |
| ② | Combine like terms: $$ 15+ \color{blue}{6i} \color{blue}{-20i} -8i^2 = -8i^2 \color{blue}{-14i} +15 $$ |
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