There are an infinite number of equations that can have 3 - i and 2 + i as its roots. If the coefficients of the root are real, the simplest equation with roots 3 - i and 2 + i is one with highest degree 4. The other two roots of the equation are 3 + i and 2 - i. The equation is:

(x - (3 - i))(x - (2 + i))(x - (3 + i))(x - (2 - i)) = 0

=> (x - 3 + i)(x - 2 - i)(x - 3 - i)(x - 2 + i) = 0

=> ((x - 3)^2 - i^2)((x - 2)^2 - i^2) = 0

=> (x^2 + 9 - 6x + 1)(x^2 - 4x + 4 + 1) = 0

=> (x^2 + 10 - 6x)(x^2 - 4x + 5) = 0

=> x^4 - 10x^3 + 39x^2 - 70x + 50 = 0

**The simplest equation with the given roots is x^4 - 10x^3 + 39x^2 - 70x + 50 = 0**

3 - i and 2 + i

(x - (3 - i)) (x - (2 + i)) (x - (3 + i)) (x - (2 - i))

(x - 3 + i) (x - 2 - i) l (x - 3 - i) (x - 2 + i)

Foil the first 2 parentheses:

x^2-2x-ix-3x + 6 + 3i + ix -2i - (-1) ( i x i = -1)

combine the terms and simplify:

x^2 - 2x - 3x - ix + ix + 3i - 2i + 6 + 1

(x^2 - 5x + i + 7) l (x - 3 - i) (x - 2 + i)

now you can either foil the result of the foil with the next parenthesis and repeat with the 4th parenthesis, or you can foil the other 2 parentheses together and then foil the result with the result of the first foil. I'll do the later:

(x - 3 - i) (x - 2 + i) should have the same answer (except the i will be negative) as the first foil because the equation is just flipped:

so:

(x^2 - 5x + i + 7) (x^2 - 5x - i + 7)

now foil:

x^4 - x^3 - ix^2 + 7x^2 - 5x^3 + 25x^2 + 5xi -35x + ix^2 - 5xi -(-1) + 7i + 7x^2 - 35x - 7i + 49

combine like terms:

x^4 - x^3 - 5x^3 - ix^2 + ix^2 + 7x^2 + 25x^2 + 7x^2 + 5xi - 5xi -35x - 35x + 7i - 7i + 49 + 1

**x^4 - 6x^3 + 39x^2 - 70x + 50**