Question
Answer
$$(4+3i)\cdot(17-6i)=-18i^2+27i+68$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4+3i)\cdot(17-6i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}68-24i+51i-18i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-18i^2+27i+68\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4+3i}\right) $ by each term in $ \left( 17-6i\right) $. $$ \left( \color{blue}{4+3i}\right) \cdot \left( 17-6i\right) = 68-24i+51i-18i^2 $$ |
| ② | Combine like terms: $$ 68 \color{blue}{-24i} + \color{blue}{51i} -18i^2 = -18i^2+ \color{blue}{27i} +68 $$ |
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