##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 :”1,1+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##,
