(3+2i)(3-2i)...Please help :)?

Answer 1

it is 9+4=13

Answer 2

Is your class teaching about how i is a special number, the square root of -1?
If so, you need to know that i^2 = -1.

Also, the following algebra shortcut is useful but not necessary:
(a + b)(a - b) = a^2 - b^2

Here is how I would solve the problem, then:
(3 + 2i)(3 - 2i)
= (3)^2 - (2i)^2
= 9 - 4i^2
= 9 - 4(-1)
= 9 + 4
= 13

If i is just a variable, and if your class isn't studying the special number i being the square root of -1, then the answer is just:
9 - 4i^2
("nine minus 4 i squared")

Answer 3

it's called square difference ... the answer is ( 3^2 - (2i)^2 )
which is ( 9 - 4*i^2 )
and since i^2 = -1
then ( 9 - 4* (-1) ) = 9+4 =13

Answer 4

(3 + 2i)(3 - 2i)
= (3)^2 - (2i)^2
= 9 - 4i^2
= 9 - 4(-1)
= 9 + 4
= 13

Answer 5

(3+2i)(3-2i)
you have to multiply 3-2i by 3 and than by 2i
9-6i+6i+4i 2(you put the 2 above the i on the right side)
-6i+6i=0 so its just
9-4i 2

Answer 6

i think this in one you use that FOIL on -- "first outside inside last"

so the first 3 times the second 3
then the first 3 times the second 2i
then the first 2i time the second 3
then the first 2i times the second 2i

3(3)+3(-2i)+2i(3)+2i(-2i) do the math and combine terms

Answer 7

Remember F.O.I.L

F = first number in each parenthesis so, 3 x 3=9
O = outer numbers in each parenthesis so, 3 x -2i = -6i
I = inner numbers in each parenthesis so, 2i x 3 = 6i
L = last numbers in each parenthesis so, 2i x -2i = -4i^2

Put it together:

9-6i+6i-4i^2=

Solve.

Answer 8

Use your FOIL rule to get 9 -4i^2.
But i^2 = -1, so the correct answer is 9 + 4 = 13.

Answer 9

has anyone ever told you about foil? first, outer,inner,last. well the answer is 9-6i+6i-4i2(squared) = 9-4i2(squared)
3x3=9, then -2x3=-6, then 2x3=6, then 2x-2=-4. the -6+6=0 and cancels out.

Answer 10

Mutliply each term in the first bracket by each term in the second bracket:
9-6i+6i-4i²

Then simplify to:
9-4i²

If you know what i is....