##x^3-3x^2+4x-2##

We know that the complex roots always occur in conjugate pairs.

One complex root is ##1+i## so there must be its conjugate i.e. ##1-i## as the other root.

Hence there are ##3 roots :11+i 1-i##.

Therefore the poly. of the least degree must be a cubic having ##3 zeroes 1 1+i and 1-i##.

Since the lead-co-eff. is ##1## the cubic poly. ##p(x)## must read :

## p(x)=(x-1)(x-(1+i))(x-(1-i))##

##=(x-1){((x-1)-i)((x-1)+i}##

##=(x-1){(x-1)^2-i^2}##

##=(x-1){(x-1)^2+1}##

##=(x-1)^3+(x-1)##

##=(x^3-1-3x^2+3x)+(x-1)##

##=x^3-3x^2+4x-2##