help on my stuck?

Question

how would you do simplify this? (imaginary #'s)

10i(2-2i) (5+i)

Answer 1

I'd start by multiplying the 10i through the first set of parentheses:

= 10i(2 - 2i)
= 20i - 20(i²)
But remember that i² = -1
= 20i - 20(-1)
= 20i + 20

Now multiply this by last equation, so your revised equation is:
(20i + 20)(5 + i)

Use the FOIL method:
First: 20i * 5 = 100i
Outer: 20i * i = 20(i²) = 20(-1) = -20
Inner: 20 * 5 = 100
Last: 20 * i = 20i

Now add them back together:
100i - 20 + 100 + 20i

Group the real and imaginary parts:
(100 - 20) + (100i + 20i)
= 80 + 120i

There's your final answer, unless you want to factor out a common 40 from each:
= 40 ( 2 + 3i )

Answer 2

imagine that i is x when you multyply by i and have in mind that i*i = -1

10i (2-2i) (5+i)=

(20i-20i*i)(5+i)=

(20i -20 (-1))(5+i)=

(20i+20)(5+i)=

100i + 20i*i +100 + 20i=

120i +20 (-1) +100=

120i +80 =

40 (3i + 2)

Answer 3

10i (2-2i) (5+i)
= 10i (0i) (5+i)
= 10i (5+i)
= 15i

Answer 4

(20i-20i) (5+i)
minus i
(20i-21i)(5)