SOLUTION: (16-5i)(-2 - i) with work shown?

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Question 775735: (16-5i)(-2 - i) with work shown?
Found 2 solutions by swincher4391, DrBeeee:
Answer by swincher4391(1107) (Show Source):
You can put this solution on YOUR website!
This is a simple case of foiling, but you have to remember that i^2 = -1.
(16-5i)(-2-i)
first : -32
outer: -16i
inner: 10i
last: 5i^2
-32 + -16i + 10i + 5i^2
-32 - 16i + 10i - 5
-6i - 37

Answer by DrBeeee(684) (Show Source):
You can put this solution on YOUR website!
Given: (16-5i)*(-2-i)
Let
(1) P = (16-5i)*(-2-i)
All you need to do is FOIL the binomials of (1) to get
(2) P = 16*(-2) + 16*(-i) -5i*(-2) -5i*(-i) or
(3) P = -32 - 16*i +10i + 5i^2 or
because i^2 = -1 we get
(4) P = -32 - 16*i +10i - 5 or
(5) P = -32 - 6i - 5 or
(6) P = -37 - 6i
Answer: (16-5i)*(-2-i) = -37 - 6i