Answer:
[tex]x^{3} +(2-3i)x^{2} -(3+6i)x+9i[/tex]
Step-by-step explanation:
- if a, b, c are the zeros of a polynomial, then the equation of the curve with leading coefficient of 1 or the polynomial function with leading coefficient of 1 f(x) can be written as : (x-a)* (x-b)* (x-c).
- here, the given zeros are : 1, -3 , 3i
- so, the polynomial function f(x) = (x-1)* (x-(-3))* (x-3i).
=(x-1)* (x+3)* (x-3i)
=[tex](x^{2} +2x-3)*(x-3i)\\=x^{3}+2x^{2} -3x-3ix^{2} -6ix+9i.\\=x^{3} +(2-3i)x^{2} -(3+6i)x+9i[/tex]
- thus, the polynomial function with the least degree and a leading coefficient of 1 that has the zeros : 1,-1(multiplicity 3), and 3i is
[tex]x^{3} +(2-3i)x^{2} -(3+6i)x+9i[/tex]
