Question
Answer
$$(3-2i)\cdot(7-5i)=10i^2-29i+21$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2i)\cdot(7-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}21-15i-14i+10i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}10i^2-29i+21\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-2i}\right) $ by each term in $ \left( 7-5i\right) $. $$ \left( \color{blue}{3-2i}\right) \cdot \left( 7-5i\right) = 21-15i-14i+10i^2 $$ |
| ② | Combine like terms: $$ 21 \color{blue}{-15i} \color{blue}{-14i} +10i^2 = 10i^2 \color{blue}{-29i} +21 $$ |
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