What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1+i?

##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##