Question
Answer
$$(4-i)\cdot(7-5i)=5i^2-27i+28$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4-i)\cdot(7-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}28-20i-7i+5i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}5i^2-27i+28\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4-i}\right) $ by each term in $ \left( 7-5i\right) $. $$ \left( \color{blue}{4-i}\right) \cdot \left( 7-5i\right) = 28-20i-7i+5i^2 $$ |
| ② | Combine like terms: $$ 28 \color{blue}{-20i} \color{blue}{-7i} +5i^2 = 5i^2 \color{blue}{-27i} +28 $$ |
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