Question
Answer
$$(2i-9)\cdot(-9-5i)=-10i^2+27i+81$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2i-9)\cdot(-9-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-18i-10i^2+81+45i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-10i^2+27i+81\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2i-9}\right) $ by each term in $ \left( -9-5i\right) $. $$ \left( \color{blue}{2i-9}\right) \cdot \left( -9-5i\right) = -18i-10i^2+81+45i $$ |
| ② | Combine like terms: $$ \color{blue}{-18i} -10i^2+81+ \color{blue}{45i} = -10i^2+ \color{blue}{27i} +81 $$ |
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