we know that
The given zeros are
x=0
x=i
x=1+i
Remember that
the complex conjugate root theorem states that if P is a polynomial in one variable with real coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a − bi is also a root of P
so
the other zeros of the given polynomial f(x) are
x=-i
x=1-i
The polynomial in factored form is given by
[tex]f(x)=x(x-i)(x+i)(x-(1+i))(x-(1-i))[/tex]
Simplify
[tex]\begin{gathered} f(x)=x(x^2-i^2)(x^2-x(1-i)-x(1+i)+(1-i^2)) \\ f(x)=x(x^2+1)(x^2-x+xi-x-xi+1+1) \\ f(x)=x(x^2+1)(x^2-2x+2) \end{gathered}[/tex]
simplify
[tex]\begin{gathered} f(x)=(x^3+x)(x^2-2x+2) \\ f(x)=x^5-2x^4+2x^3+x^3-2x^2+2x \\ f(x)=x^5-2x^4+3x^3-2x^2+2x \end{gathered}[/tex]
