What is the result of the difference (2+i)(3-2i)-(1-2i)(2-i) ?

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We'll use the property of multiplication to be distributive over addition:

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

(2+i)(3-2i) = 6 - 4i + 3i - 2i^2

(2+i)(3-2i) = 6 - i - 2i^2

But i^2 = -1:

(2+i)(3-2i) = 6 - i + 2

(2+i)(3-2i) = 8 - i (1)

We'll calculate the 2nd product:

(1-2i)(2-i) = 1*(2-i) - 2i*(2-i)

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

(1-2i)(2-i) = 2 - 5i - 2

(1-2i)(2-i) = -5i (2)

We'll subtract (2) from (1):

8 - i - (5i) = 8 - i + 5i = 8 + 4i

**The result of difference is: (2+i)(3-2i) - (1-2i)(2-i) = 8 + 4i.**

We need to find the value of (2+i)(3-2i)-(1-2i)(2-i)

(2+i)(3-2i) - (1-2i)(2-i)

open the brackets and multiply

=> 6 - 4i + 3i - 2i^2 - [ 2 - i - 4i + 2i^2]

we know i^2 = -1

=> 6 - 4i + 3i + 2 - 2 + i + 4i + 2

=> 8 + 4i

**The required result is 8 + 4i**

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