Question

A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i what is the students mistakes

Answer 1

1. In the multiplication of the imaginary parts, the student forgot to square of i. OR

2. The student has only multiplied the real parts and the imaginary parts.

Correct value
`=22+7i`.

Explanation:

The given expression is

`(4+5i)(3-2i)`

A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i.

Student's mistake can be either 1 or second:

1. In the multiplication of the imaginary parts, the student forgot to square of i.

2. The student has only multiplied the real parts and the imaginary parts.

`(4+5i)(3-2i)=4* 3+5* (-2)i=12-10i`

Which is not correct. The correct steps are shown below.

Using distributive property, we get

`4(3-2i)+5i(3-2i)`

`4(3)+4(-2i)+5i(3)+5i(-2i)`

`12-8i+15i-10i^2`

`12+7i-10(-1)`
`[\because i^2=-1]`

`12+7i+10`

`22+7i`

Therefore, the correct value of
`(4+5i)(3-2i)` is
`22+7i`.