Question
Answer
$$(3-4i)\cdot(6-5i)=20i^2-39i+18$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-4i)\cdot(6-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}18-15i-24i+20i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}20i^2-39i+18\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-4i}\right) $ by each term in $ \left( 6-5i\right) $. $$ \left( \color{blue}{3-4i}\right) \cdot \left( 6-5i\right) = 18-15i-24i+20i^2 $$ |
| ② | Combine like terms: $$ 18 \color{blue}{-15i} \color{blue}{-24i} +20i^2 = 20i^2 \color{blue}{-39i} +18 $$ |
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