Question 238136
two roots of this equation are:


x = 3
x = 2+i


x - 3 = 0 

or


x - 2 - i = 0




it seems you're on the right track with the conjugate of x = 2-i


that should make them cancel out when you multiply them together.


let's do that and see how it works.


x = 2-i means that x - 2 + i = 0


you have 3 factors.


they are:


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


first you want to multiply x - 2 + i by x - 2 - i


each term in one factor has to be multiplied by each term in the other factor.


start with x in the left factor and multiply each term in the right factor from left to right.


x * x = x^2
x * -2 = -2x
x * -i = -ix


do -2 next.


-2 * x = -2x
-2 * -2 = 4
-2 * -i = 2i


do i next.


i * x = ix
i * -2 = -2i
i * -i = -i^2 = 1 because i^2 = -1


add all your terms together to get:


x^2 - 2x - ix -2x + 4 + 2i + ix -2i + 1


+ ix and - ix cancel out
+ 2i and - 2i cancel out
-2x and -2x become -4x
+ 4 and + 1 become + 5


your equation becomes x^2 - 4x + 5


multiply this equation by x-3 to get:


x * x^2 = x^3
x * -4x = -4x^2
x * 5 = 5x
-3 * x^2 = -3x^2
-3 * -4x = 12x
-3 * 5 = -15


add these terms together to get:


x^3 - 4x^2 + 5x - 3x^2 + 12x - 15


combine like terms to get:


x^3 -7x^2 + 17x - 15


that's your equation.


a graph of this equation looks like:


{{{graph(400,400,-20,20,-20,20,x^3 - 7x^2 + 17x - 15)}}}


It only has one real root (x-axis crossing point).


If you set x = 3, the equation should become equal to 0.


I confirmed that it does.


If you set x = 2-i, the equation should become equal to 0.


I confirmed that it does.  It was a nosebleed but it was confirmed.


I suspect 2+i will confirm as well although I'm not going to go through that again.


Not today, anyway.