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

1 Answer
Apr 12, 2016

-112+92i112+92i

Explanation:

Multiply the two binomials by the FOIL method

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

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

-120+12i+80i-8i^2120+12i+80i8i2
since i^2 = -1i2=1 from using Euler's notation
i=sqrt(-1)i=1
i^2 = (sqrt(-1))(sqrt(-1)) = -1i2=(1)(1)=1

we get

-120+12i+80i-8(-1)120+12i+80i8(1)
-120+12i+80i+8120+12i+80i+8

now combine like terms
-112+92i112+92i