$3 i \cdot (2 - 5 i) \cdot (4 + i) = 6 i^{2} + 39 i + 60$

Explanation

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$3 i \cdot (2 - 5 i) \cdot (4 + i) \overset{◯}{1} (6 i - 15 i^{2}) \cdot (4 + i) \overset{◯}{2} (6 i + 15) \cdot (4 + i) \overset{◯}{3} 24 i + 6 i^{2} + 60 + 15 i \overset{◯}{4} 6 i^{2} + 39 i + 60$
①	Multiply $3 i$ by $(2 - 5 i)$ $3 i \cdot (2 - 5 i) = 6 i - 15 i^{2}$
②	$- 15 i^{2} = - 15 \cdot (- 1) = 15$
③	Multiply each term of $(6 i + 15)$ by each term in $(4 + i)$ . $(6 i + 15) \cdot (4 + i) = 24 i + 6 i^{2} + 60 + 15 i$
④	Combine like terms: $24 i + 6 i^{2} + 60 + 15 i = 6 i^{2} + 39 i + 60$

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3i⋅(2−5i)⋅(4+i)=6i2+39i+60

$3 i \cdot (2 - 5 i) \cdot (4 + i) = 6 i^{2} + 39 i + 60$