Question
Answer
$$3i\cdot(2-i)\cdot(4+2i)=12i^2+30i+12$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3i\cdot(2-i)\cdot(4+2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(6i-3i^2)\cdot(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(6i+3)\cdot(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}24i+12i^2+12+6i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}12i^2+30i+12\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{3i} $ by $ \left( 2-i\right) $ $$ \color{blue}{3i} \cdot \left( 2-i\right) = 6i-3i^2 $$ |
| ② | $$ -3i^2 = -3 \cdot (-1) = 3 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{6i+3}\right) $ by each term in $ \left( 4+2i\right) $. $$ \left( \color{blue}{6i+3}\right) \cdot \left( 4+2i\right) = 24i+12i^2+12+6i $$ |
| ④ | Combine like terms: $$ \color{blue}{24i} +12i^2+12+ \color{blue}{6i} = 12i^2+ \color{blue}{30i} +12 $$ |
This page was created using
Complex Expression calculator