Write a polynomial function ff of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. write the polynomial in standard form. 3, 4+2i, 1+7√3, 4+2i, 1+7

I am answering based on the assumption that the last zero is 1 + 7√3

FIRST, set up the zeros as factors:
x=3 --> (x - 3)=0
x=4+2i  --> [x - (4 + 2i)] = 0
don't forget the conjugate! (which is x=4-2i) --> [x - (4 - 2i)] = 0
x=1+7√3 --> [x - (1+7√3)] = 0
don't forget to rationalize it! (x=1-7√3)  --> [x - (1-7√3)] = 0 
x=4+2i was already given as a zero so this factor and its conjugate have a multiplicity of 2 (meaning it has an exponent of 2)
x=1+7√3 was also previously given as a zero so it also has a multiplicity of 2.

NEXT, multiply all of the factors:
(x-3) x [[x-(4+2i)][x-(4-2i)]]²  x [[x-(1+7√3)][x-(1-7√3)]]² = 0
--> (x - 3) x (x² - 8x +20)² x (x² - 2x - 146)² = 0
--> x^9 - 23x^8 - 60x^7 + 4816x^6 - 29668x^5 - 140860x^4 + 2484064x^3 - 12331872x^2 + 28288960x - 25579200 = 0
--> f(x) = x^9 - 23x^8 - 60x^7 + 4816x^6 - 29668x^5 - 140860x^4 + 2484064x^3 - 12331872x^2 + 28288960x - 25579200

 

Bis your answer. it only holds the decimal numerals