Question
Answer
$$(i-2)(2\cdot(1+i)-3(i-1))=-i^2+7i-10$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(i-2)(2\cdot(1+i)-3(i-1))& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(i-2)(2+2i-(3i-3)) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(i-2)(2+2i-3i+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(i-2)(-i+5) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-i^2+5i+2i-10 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-i^2+7i-10\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2} $ by $ \left( 1+i\right) $ $$ \color{blue}{2} \cdot \left( 1+i\right) = 2+2i $$Multiply $ \color{blue}{3} $ by $ \left( i-1\right) $ $$ \color{blue}{3} \cdot \left( i-1\right) = 3i-3 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3i-3 \right) = -3i+3 $$ |
| ③ | Combine like terms: $$ \color{blue}{2} + \color{red}{2i} \color{red}{-3i} + \color{blue}{3} = \color{red}{-i} + \color{blue}{5} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{i-2}\right) $ by each term in $ \left( -i+5\right) $. $$ \left( \color{blue}{i-2}\right) \cdot \left( -i+5\right) = -i^2+5i+2i-10 $$ |
| ⑤ | Combine like terms: $$ -i^2+ \color{blue}{5i} + \color{blue}{2i} -10 = -i^2+ \color{blue}{7i} -10 $$ |
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