How do you simplify #(-12+8i)(10-i) #?

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Answer 1 · 2016-04-12T15:35:12

#-112+92i#

Multiply the two binomials by the FOIL method

#(-12+8i)(10-i)#

Firsts #(-12)(10) = -120#
Outers #(-12)(-i) = 12i#
Inners #(8i)(10) = 80i#
Lasts (8i)(-i) = -8i^2#

#-120+12i+80i-8i^2#
since #i^2 = -1# from using Euler's notation
#i=sqrt(-1)#
#i^2 = (sqrt(-1))(sqrt(-1)) = -1#

we get

#-120+12i+80i-8(-1)#
#-120+12i+80i+8#

now combine like terms
#-112+92i#