(3i+5)(2-i)

Question

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

Answer 1

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

 

 

Answer 2

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

Answer 3

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