How would you expand simplify this has i?

Question

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

Answer 1

i stands for the square root of  negative one to let you know.

its an imaginary number as there never can be such a thing as i in the real world but its useful in finding roots for equations such as this. (i.e. going backwards essentially in this problem taking what would be your answer and trying to find this )

Anyways we should deal with the is first.

(x-2)(x-3-i)(x-3+i) ; distribute the trinomials

(x-2)(x^2-3x+ix-3x+9-3i-ix+3i+1) ; the next step is to collect like terms

(x-2)(x^2-6x+10) ; Now notice that in circumstances such as you had in the originals trinomials that all of the is will eventually cancel out because its a conjugate. Also if you draw x^2-6x+10 or anything that has all imaginary/complex roots then it will never cross the x axis, if it does then something is wrong

now finish distributing and simplifying;

(x^3-6x^2+10x-2x^2+12x-20)

x^3-8x^2+22x-20

Answer 2

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

(x-2)[(x-3)^2-i^2]=

(x-2)(x^2-6x+10)=

x^3-8x^2+22x-20