Question 1201671
P(x) =2x^3-5x^2+6x-2; 1+i
There have to be three factors

1-i is the conjugate of 1+i is also a root

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

=x^2 -x +xi -x-xi+1+1

= x^2 -2x +2

{{{(2x^3-5x^2+6x-2)/(x^2-2x+2)}}}  You can divide by co-efficient form
= 2x-1

2x-1=0
x=1/2
The factors are(1+i),(1-i),(1/2)