Answer:
[tex]a. \ (x-2)(x+i)(x-i)[/tex]
Step-by-step explanation:
First of all let's notice that [tex](x-i)(x-i)=(x-i)^2 = x^2+2xi-1 \in \mathbb{C}[/tex] while our original polynomial is real. So we can rule out the even options B and D. At this point, if the polynomial has a factor of [tex](x-\alpha)[/tex] it means tha [tex]\alpha[/tex] is a zero of the polynomial. Let's check both 2 (for option a) and -2 (for option c)
[tex]p(2)=2^3-2(2)^2+2-2= 8-8+2-2=0[/tex]
[tex]p(-2)= (-2)^3-2(-2)^2+(-2)-2=-8-8-2-2=-20[/tex]
At this point 2 is a zero, and our final factoring is a[tex](x-2)(x+i)(x-i)[/tex]
