How do you simplify #(-2+2i)(-1-i)(3-3i)#?

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Answer 1 · 2017-04-19T21:32:55

Use the F.O.I.L. method.

Given: #(-2+2i)(-1-i)(3-3i) = ?#

Let's use the F.O.I.L method on the first pair:

#(-2+2i)(-1-i)=(-2)(-1)+ (-2)(-i)+(2i)(-1)+(2i)(-i)#

#(-2+2i)(-1-i)=2+ 2i-2i-2i^2#

#(-2+2i)(-1-i)=2-2i^2#

Use #i^2=-1#:

#(-2+2i)(-1-i)=2+2#

#(-2+2i)(-1-i)=4#

Multiply the above by the 3rd factor:

#(-2+2i)(-1-i)(3-3i) = 4(3-3i)#

#(-2+2i)(-1-i)(3-3i) = 12-12i#