Answer
$$(3ix+9)(ix-3)=3i^2x^2-27$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3ix+9)(ix-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3i^2x^2-9ix+9ix-27 \xlongequal{ } \\[1 em] & \xlongequal{ }3i^2x^2 -\cancel{9ix}+ \cancel{9ix}-27 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3i^2x^2-27\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3ix+9}\right) $ by each term in $ \left( ix-3\right) $. $$ \left( \color{blue}{3ix+9}\right) \cdot \left( ix-3\right) = 3i^2x^2 -\cancel{9ix}+ \cancel{9ix}-27 $$ |
| ② | Combine like terms: $$ 3i^2x^2 \, \color{blue}{ -\cancel{9ix}} \,+ \, \color{blue}{ \cancel{9ix}} \,-27 = 3i^2x^2-27 $$ |
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