Question
Answer
$$(2+3i)\cdot(1-4i)=-12i^2-5i+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2+3i)\cdot(1-4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2-8i+3i-12i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-12i^2-5i+2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2+3i}\right) $ by each term in $ \left( 1-4i\right) $. $$ \left( \color{blue}{2+3i}\right) \cdot \left( 1-4i\right) = 2-8i+3i-12i^2 $$ |
| ② | Combine like terms: $$ 2 \color{blue}{-8i} + \color{blue}{3i} -12i^2 = -12i^2 \color{blue}{-5i} +2 $$ |
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