How do you simplify #(3-5i)(4+6i)#?

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Answer 1 · 2018-08-08T01:37:49

#42 - 2i#

#(3-5i)(4+6i)#

First multiply the first values:
#3 * 4 = 12#

Outer values:
#3 * 6i = 18i#

Inner values:
#-5i * 4 = -20i#

Last values:
#-5i * 6i = -30i^2#

Combine them all together:
#12 + 18i - 20i - 30i^2#

Combine like terms:
#12 - 2i - 30i^2#

We know that #i^2# is equivalent to #-1#, so:
#12 - 2i - 30(-1)#

Simplify:
#12 - 2i + 30#

#42 - 2i#

Hope this helps!