How do you multiply #3i(4+2i)(2+5i)#?

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Answer 1 · 2017-02-09T16:13:03

The answer is #=-72-6i#

Remember that

#i^2=-1#

Therefore,

#(4+2i)(2+5i)=8+20i+4i+10i^2#

#=8-10+24i#

#=-2+24i#

so,

#3i(4+2i)(2+5i)=3i(-2+24i)#

#=-6i+72i^2#

#=-72-6i#