How do you multiply #(3-2i)(5-6i)#?

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Answer 1 · 2016-01-14T07:19:30

Steps are given below.

#(3-2i)(5-6i)#
Use FOIL

#=(3)(5) + (3)(-6i) - (2i)(5) + (-2i)(-6i)#

#=15 - 18i - 10i + 12i^2#

#color(red) "note : i² =-1"#

#=15 -28i + 12(-1)#

#=15-28i-12#

#=3-28i#