How do you simplify #6i(3+2i)#?

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Answer 1 · 2016-09-16T20:17:36

#- 12 + 18 i#

We have: #6 i (3 + 2 i)#

Let's expand the parentheses:

#= (6 i) (3) + (6 i) (2 i)#

#= 18 i + 12 i^(2)#

Let's apply the fact that #i^(2) = - 1#:

#= 18 i + 12 cdot - 1#

#= - 12 + 18 i#