Answer:
[tex]f(x)=3x^3+x^2+3x+1[/tex]
Step-by-step explanation:
If a real number [tex]-\frac{1}{3}[/tex] is a zero of polynomial function, then
[tex]x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}[/tex]
is the factor of this function.
If a complex number [tex]-i[/tex] is a xero of the polynomial function, then the complex number [tex]i[/tex] is also a zero of this function and
[tex]x-(-i)=x+i\ \text{ and }\ x-i[/tex]
are two factors of this function.
So, the function of least degree is
[tex]f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)[/tex]
If the polynomial function must be with integer coefficients, then it has a form
[tex]f(x)=3x^3+x^2+3x+1[/tex]
