Question 1126012

if zeros are {{{x[1]=3}}} (real one)  and {{{x[2]=2 + i}}} (complex one), then you also have {{{x[3]=2 - i}}}  because complex zeros always come in pairs

since you have three zeros, a polynomial function of least degree will be a polynomial function of  degree {{{3}}}

{{{f(x)=(x-x[1])(x-x[2])(x-x[3])}}}...substitute given zeros

{{{f(x)=(x-3)(x-(2 + i))(x-(2 - i))}}}
{{{f(x)=(x-3)(x-2 -i)(x-2 + i)}}}
{{{f(x)=(x-3)(x^2 - 4 x + 5)}}}
{{{f(x)=x^3 - 7 x^2 + 17 x - 15}}}