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## (3i+5)(2-i)

I just don't know how to solve this problem. I know the answer is 13+i.

The product is
= 6i + 10 -5i - 3i.i where the imaginary value i = sqrt (i) therefore i.i = -1
= 6i - 5i + 10 - 3(-1)
= i +10 + 3
= 13 + i

Hi Shyanna;
I RESEARCHED IT.  AN IMAGINARY NUMBER MULTIPLIED BY AN IMAGINARY NUMBER RESULTS IN A NEGATIVE NUMBER.  HOWEVER, HEREIN, WE ARE MULTIPLYING (3i)(-i).

(3i+5)(2-i)
Let's FOIL...
FIRST...(3i)(2)=6i
OUTER...(3i)(-i)=--3=3
INNER...(5)(2)=10
LAST...(5)(-i)=-5i
6i+3+10-5i
i+13

Hello Vivian:

(i) ( i) = -1 /  here does not result in a positive number.

( i ) ( -i) = - ( -1 ) = 1

i ^3 = ( i^2) ( i) = -i

i ^4 = ( i ^2) ( i^2) = ( -1) ( -1) = 1

Imaginary numbers raised to the power of numbers greater than 4 :

i ^ n  for n> 4

i ^ n = i ^ ( 4m + r)

for  r =1

i ^ ( 4m +r ) = i ^(4m) . i = i     i.e. i ^65 = i^64. i = i

for  r =2
i ^ ( 4m + r)  = i ^ 4m . i ^2 =  i^2 = 1    i,e   i^10 = i( 8 +2) = i^8 . i^2 = -1

for  r = 3

i ^ ( 4m +3 ) = i ^4m . i ^3 =  i^3 = - i     i.e.  i ^27 = i ^( 24 +3) = i ^24 . i^3 = -i

i is defined to be sqrt(-1) and is called the imaginary unit.

i^2 = -1
i^3 = i^2 * i = -i
i^4 = i^3 * i = -i * i = - i^2 = - - 1 = 1
i^5 = i^4 * i = 1 * i = i
i^6 = i^5 * i = i * i = i^2 = -1
etc ...

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

First use the distributive property:
3i(2-i) +5(2-i) =
6i - 3i^2 +10 - 5i =

Then commutative property of addition and i^2 = -1:
6i - 5i - 3(-1) + 10 =
i + 3 + 10 =
13 + i