Question
Answer
$$(i^3+2)(i-2)(i^2-3)=-16i+12$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(i^3+2)(i-2)(i^2-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(-i+2)(i-2)\cdot(-1-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(-i+2)(i-2)\cdot(-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(-i^2+2i+2i-4)\cdot(-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(-i^2+4i-4)\cdot(-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}(1+4i-4)\cdot(-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}(4i-3)\cdot(-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}-16i+12\end{aligned} $$ | |
| ① | $$ i^3 = \color{blue}{i^2} \cdot i =
( \color{blue}{-1}) \cdot i =
- \, i $$$$ i^2 = -1 $$ |
| ② | Combine like terms: $$ \color{blue}{-1} \color{blue}{-3} = \color{blue}{-4} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{-i+2}\right) $ by each term in $ \left( i-2\right) $. $$ \left( \color{blue}{-i+2}\right) \cdot \left( i-2\right) = -i^2+2i+2i-4 $$ |
| ④ | Combine like terms: $$ -i^2+ \color{blue}{2i} + \color{blue}{2i} -4 = -i^2+ \color{blue}{4i} -4 $$ |
| ⑤ | $$ -i^2 = -(-1) = 1 $$ |
| ⑥ | Combine like terms: $$ \color{blue}{1} +4i \color{blue}{-4} = 4i \color{blue}{-3} $$ |
| ⑦ | $$ \left( \color{blue}{4i-3}\right) \cdot -4 = -16i+12 $$ |
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