##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##,

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