Question
Answer
$$i\cdot(3-i)\cdot(4+2i)=6i^2+14i+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}i\cdot(3-i)\cdot(4+2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(3i-i^2)\cdot(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(3i+1)\cdot(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}12i+6i^2+4+2i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}6i^2+14i+4\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{i} $ by $ \left( 3-i\right) $ $$ \color{blue}{i} \cdot \left( 3-i\right) = 3i-i^2 $$ |
| ② | $$ -i^2 = -(-1) = 1 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{3i+1}\right) $ by each term in $ \left( 4+2i\right) $. $$ \left( \color{blue}{3i+1}\right) \cdot \left( 4+2i\right) = 12i+6i^2+4+2i $$ |
| ④ | Combine like terms: $$ \color{blue}{12i} +6i^2+4+ \color{blue}{2i} = 6i^2+ \color{blue}{14i} +4 $$ |
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