SOLUTION: Express the product of 4 – 3i and 2 + i in simplest a + bi form. 1. 2 + 11i 2. 2 – 11i 3. 11 + 2i 4. 11 – 2i I have an application on my graphing calcu

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Express the product of 4 – 3i and 2 + i in simplest a + bi form. 1. 2 + 11i 2. 2 – 11i 3. 11 + 2i 4. 11 – 2i I have an application on my graphing calcu      Log On


   

Question 253915: Express the product of 4 – 3i and 2 + i in simplest a + bi form.
1. 2 + 11i
2. 2 – 11i
3. 11 + 2i
4. 11 – 2i
I have an application on my graphing calculator allowing me to add/subtract/multiply/or divide complex numbers. I used that and came up with 4-5i? Yet that isn't an answer choice?
Found 3 solutions by Theo, scott8148, richwmiller:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i came up with 11 - 2i

here's how.

multiply (4-3i) * (2+i)

4 * 2 = 8
4 * i = 4i
-3i * 2 = -6i
-3i * i = -3i^2

add them up together and you get:


8 + 4i - 6i - 3i^2

combine like terms to get:

8 - 2i - 3i^2

since i^2 = -1, this becomes:

8 - 2i - 3*(-1) which becomes:

8 - 2i + 3

combine like terms to get:

11 - 2i

that would be selection 4.





Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
FOILing the binomials ___ 8 + 4i - 6i - 3i^2 ___ 8 - 2i + 3 = 11 - 2i

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
I get 11-2i.
foil
(4 – 3i) (2 + i)
4*2-3i*2 +4i -3i^2=
8-6i+4i+3=
11-2i
Did you multiply or subtract or add?
I even tried add, subtract and divide and still don't get your 4-5i
Maybe you had the signs wrong.