Given the zeros:
[tex]\begin{gathered} x_1=4 \\ x_2=4-5i \end{gathered}[/tex]You need to remember that the Factor Theorem states that, if, for a polynomial:
[tex]f(a)=0[/tex]Then, this is a factor of the polynomial:
[tex](x-a)[/tex]In this case, you know that:
[tex]\begin{gathered} f(4)=0 \\ f(4-5i)=0 \end{gathered}[/tex]Therefore, you can determine that these are factors of the polynomial:
[tex]\begin{gathered} (x-4) \\ (x-(4-5i)) \end{gathered}[/tex]By definition, Complex Conjugates have this form:
[tex](a+bi)(a-bi)[/tex]Therefore, you can determine that this is also a factor:
[tex](x-(4+5i))[/tex]Now you can set up that the Factored Form of the polynomial is:
[tex](x-4)(x-(4-5i))(x-(4+5i))[/tex]You need to expand the expression by applying the Distributive Property and applying:
[tex](a-b)(a+b)=a^2-b^2[/tex]Then:
[tex]=(x-4)((x-4)^2-25i^2)[/tex]By definition:
[tex](a-b)^2=a^2-2ab+b^2[/tex]Then, you get:
[tex]=(x-4)(x^2-(2)(x)(4)+4^2-25i^2)[/tex][tex]=(x-4)(x^2-8x+16-25i^2)[/tex]Knowing that:
[tex]i^2=-1[/tex]And adding the like terms, you get:
[tex]=(x-4)(x^2-8x+16-25(-1))[/tex][tex]=(x-4)(x^2-8x+16+25)[/tex][tex]=(x-4)(x^2-8x+41)[/tex]Applying the Distributive Property and adding the like terms, you get:
[tex]=(x^2)(x)-(x)(8x)+(x)(41)-(4)(x^2)+(4)(8x)-(4)(41)[/tex][tex]=x^3-8x^2+41x-4x^2+32x-164[/tex][tex]=x^3-12x^2+73x-164[/tex]Hence, the answer is: Second option.