Answer:
[tex]f(x)=x^{3}-3x^{2} +4x-2[/tex]
Step-by-step explanation:
we know that
The conjugate root theorem states that if the complex number a + bi is a root of a polynomial P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of that polynomial
In this problem we have that
The polynomial has roots 1 and (1+i)
so
by the conjugate root theorem
(1-i) is also a root of the polynomial
therefore
The lowest degree of the polynomial is 3
so
[tex]f(x)=a(x-1)(x-(1+i))(x-(1-i))[/tex]
Remember that
The leading coefficient is 1
so
a=1
[tex]f(x)=(x-1)(x-(1+i))(x-(1-i))\\\\f(x)=(x-1)[x^{2} -(1-i)x-(1+i)x+(1-i^2)]\\\\f(x)=(x-1)[x^{2} -x+xi-x-xi+2]\\\\f(x)=(x-1)[x^{2} -2x+2]\\\\f(x)=x^{3}-2x^{2} +2x-x^{2} +2x-2\\\\f(x)=x^{3}-3x^{2} +4x-2[/tex]
