Answer
$$3i\cdot(4-5i)-1=12i+15-1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3i\cdot(4-5i)-1& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12i-15i^2-1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12i+15-1\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{3i} $ by $ \left( 4-5i\right) $ $$ \color{blue}{3i} \cdot \left( 4-5i\right) = 12i-15i^2 $$ |
| ② | $$ -15i^2 = -15 \cdot (-1) = 15 $$ |
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