Question
Answer
$$-4\cdot(2-i)\cdot(15+5i)=20i-140$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-4\cdot(2-i)\cdot(15+5i)& \xlongequal{ }-(8-4i)\cdot(15+5i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(120+40i-60i-20i^2) \xlongequal{ } \\[1 em] & \xlongequal{ }-(-20i^2-20i+120) \xlongequal{ } \\[1 em] & \xlongequal{ }-(20-20i+120) \xlongequal{ } \\[1 em] & \xlongequal{ }-(-20i+140) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}20i-140\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{8-4i}\right) $ by each term in $ \left( 15+5i\right) $. $$ \left( \color{blue}{8-4i}\right) \cdot \left( 15+5i\right) = 120+40i-60i-20i^2 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(-20i+140 \right) = 20i-140 $$ |
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