imaginary numbers?

Question

I can't figure them out Help if you can. It says simplify:
(i-7)(3i+3) and -7-10i-(-15-3i).

Answer 1

(i - 7)(3i + 3) =

3i² - 18i - 21

-1(3) - 18i - 21

- 3 - 18i - 21

18i - 24

- - - - - - - - - -s-

Answer 2

<< remember that ii = i * i = i^2 = -1 >>

(i-7)(3i+3) = (3ii + 3i) - (21i + 21)
= (3i -3) - (21i + 21)
= -18i -24 = 6(-3i - 4)

-7-10i-(-15-3i) = -7 - 10i + (-1) * (-15 - 3i) = -7 - 10i + 15 + 3i
= 8 - 7i

don't be afraid of the the name "imaginary numbers" or "complex numbers". they could just as well be called "blue numbers"

Answer 3

Multiplying complex numbers is done the same way as multiplying binomials. All the usual rules of algebra apply. Once you've obtained all the terms, then collect the real parts and the imaginary parts separately. Keep in mind that i^2 = -1

3i+3
i-7
--------- multiply
-3+3i
-21i -21
------------ add
-24+i

For the second problem, it's just to make sure that you remember the distributive law correctly. Keep in mind that subtracting a number is the same as adding its opposite.

Answer 4

I hope this is correct for the first equation:

(i-7)(3i + 3)
3i2(squared)+3i-21i-21
3i2(sqaured)-18i-21

Answer 5

just remember i*i = -1

(i-7) * (3i + 3) (foil method)

(3i*i) + (i*3) + (-7*3i) + (-7*3)
(-3) + 3i - 21i -21
i(3-21) -24
-18i-24

Answer 6

Do the math, just like you would if 'X' was in the problem.
Then change each of the i^2 to -1.

For example: (i - 9)(2i + 5)
= 2i^2 + 5i -18i - 45
= 2i^2 + 23i - 45
= 2(-1) + 23i - 45
= -2 + 23i -45
= -47 + 23i

Answer 7

666

Answer 8

twelvty