How do you solve #(-3i)(2i)(8+2i)#?

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Answer 1 · 2018-01-19T15:25:46

Multiply them together as if #i# is a variable, and then evaluate for #i# using #i^2 = -1# to get #(-3i)(2i)(8 + 2i) = 48 + 12i#.

Well, we can simply multiply them as we would with the reals, first treating #i# as a variable, and then evaluate for its value later on.

#(-3i)(2i)(8 + 2i)#

We can start with #(-3i)(2i)#, which becomes #-6i^2#:

#= (-6i^2)(8 + 2i)#

Then use the distributive property to "multiply over" the first term to the second term:

#= 8(-6i^2) + 2i(-6i^2)#

#= -48i^2 + -12i^3#

Now we can start evaluating. Let's first "split" #i^3# into #i^2# and #i#:

#= -48i^2 + -12 i^2 i#

We know that by definition, #i^2 = -1#:

#= -48(-1) + -12(-1)i#

#= 48 + 12i#

Ah, that's as far as we can go! Therefore,

#(-3i)(2i)(8 + 2i) = 48 + 12i#.