Question
Answer
$$(6+8i)\cdot(3-4i)=-32i^2+18$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6+8i)\cdot(3-4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}18-24i+24i-32i^2 \xlongequal{ } \\[1 em] & \xlongequal{ }18 -\cancel{24i}+ \cancel{24i}-32i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-32i^2+18\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{6+8i}\right) $ by each term in $ \left( 3-4i\right) $. $$ \left( \color{blue}{6+8i}\right) \cdot \left( 3-4i\right) = 18 -\cancel{24i}+ \cancel{24i}-32i^2 $$ |
| ② | Combine like terms: $$ 18 \, \color{blue}{ -\cancel{24i}} \,+ \, \color{blue}{ \cancel{24i}} \,-32i^2 = -32i^2+18 $$ |
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