Answer
$$(-4-i)\cdot(2-4i)=4i^2+14i-8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-4-i)\cdot(2-4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-8+16i-2i+4i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4i^2+14i-8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{-4-i}\right) $ by each term in $ \left( 2-4i\right) $. $$ \left( \color{blue}{-4-i}\right) \cdot \left( 2-4i\right) = -8+16i-2i+4i^2 $$ |
| ② | Combine like terms: $$ -8+ \color{blue}{16i} \color{blue}{-2i} +4i^2 = 4i^2+ \color{blue}{14i} -8 $$ |
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