How do you multiply #-2i(3-7i)(4+2i)#?

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Answer 1 · 2016-09-04T16:50:31

#(-2i)(3-7i)(4+2i)=-44-52i#

While multiplying complex numbers we should always remember that #i^2=-1#.

Hence, #(-2i)(3-7i)(4+2i)#

= #((-2i×3)-2i×(-7i))(4+2i)#

= #(-6i+14i^2)(4+2i)#

= #(-6i+14×(-1))(4+2i)#

= #(-14-6i)(4+2i)#

= #-14×4-14×2i-6i×4-6i×2i#

= #-56-28i-24i-12×(-1)#

= #-56-28i-24i+12#

= #-44-52i#