Question
Answer
$$-3i\cdot(-4+4i)+5i\cdot(3-2i)=27i+22$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-3i\cdot(-4+4i)+5i\cdot(3-2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(-12i+12i^2)+15i-10i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-(-12i-12)+15i+10 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}12i+12+15i+10 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}27i+22\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{3i} $ by $ \left( -4+4i\right) $ $$ \color{blue}{3i} \cdot \left( -4+4i\right) = -12i+12i^2 $$Multiply $ \color{blue}{5i} $ by $ \left( 3-2i\right) $ $$ \color{blue}{5i} \cdot \left( 3-2i\right) = 15i-10i^2 $$ |
| ② | $$ -10i^2 = -10 \cdot (-1) = 10 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left(-12i-12 \right) = 12i+12 $$ |
| ④ | Combine like terms: $$ \color{blue}{12i} + \color{red}{12} + \color{blue}{15i} + \color{red}{10} = \color{blue}{27i} + \color{red}{22} $$ |
This page was created using
Complex Expression calculator