Compare the quantity in Column A with the quantity in Column B.?

Question

Given the information:

a+bi=(3+i)(4-2i)

c+di=(-2-3i)(3+5i)

Column A Column B

b d

a. the quantity in Column A is greater

b. the quantity in Column B is greater

c. the two quantities are equal

d. the relationship cannot be determined on the basis of the information supplied

Answer 1

a + bi = (3 + i)(4 - 2i)

-Use FOIL: 12-6i+4i-2i^2
-Since i^2=-1, turn it to -2.  Therefore, it's: 12-6i+4i+2 (or minus -2)
-Put together like terms: 12+2; -6i+4i--You get 14-10i.
14-10i = a+bi.

c + di = (-2 - 3i)(3 + 5i)
-Use FOIL: -6-10i-9i-15i^2.
-Since i^2=-1, turn it to -15. Therefore, it's: -6-10i-9i+15 (or minus -15)
-Put together like terms: -6+15; -10i-9i--You get 9-19i.
9-19i= c+di.

Now compare Column A, which will be "b", to Column B, which will be "d".
AKA-- compare -10 to -19.

I really hope it helps!
Is this equation from Keystone's Algebra II Exam 3? Yeahh, I'm doing that now :P

Answer 2

a + bi = (3 + i)(4 - 2i)

-Use FOIL: 12-6i+4i-2i^2
-Since i^2=-1, turn it to -2.  Therefore, it's: 12-6i+4i+2 (or minus -2)
-Put together like terms: 12+2; -6i+4i--You get 14-10i.
14-10i = a+bi.

c + di = (-2 - 3i)(3 + 5i)
-Use FOIL: -6-10i-9i-15i^2.
-Since i^2=-1, turn it to -15. Therefore, it's: -6-10i-9i+15 (or minus -15)
-Put together like terms: -6+15; -10i-9i--You get 9-19i.
9-19i= c+di.

Now compare Column A, which will be "b", to Column B, which will be "d".
AKA-- compare -10 to -19.

I really hope it helps!
Is this equation from Keystone's Algebra II Exam 3? Yeahh, I'm doing that now :P