Answer:
![f(x)=x^5-x^4+13x^3-13x^2+36x-36](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9co5umi0plkf6qdimgzbwqfpyaaxulwfar.png)
Explanation:
If the given zeros are 2i, 3i, and 1, the zeros of the function are: 2i, 3i, 1, -2i, and -3i, then the polynomial function is:
![f(x)=(x-2i)(x+2i)(x-3i)(x+3i)(x-1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rbtpu86aewpho6iel671he6wxs19t6bv3z.png)
Using that the product of the difference and the sum is equal to the difference of the squares:
![(a-b)(a+b)=a^2-b^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/v8av7er5qw0zq0pqq3mqb9swutp0hfyssv.png)
![f(x)=(x^2-(2i)^2)(x^2-(3i)^2)(x-1)\\ f(x)=(x^2-2^2i^2)(x^2-3^2i^2)(x-1)\\ i^2=-1\\ f(x)=(x^2-4(-1))(x^2-9(-1))(x-1)\\ f(x)=(x^2+4)(x^2+9)(x-1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2dv9pkuxl4703owid4yu3e0xzxwp3q1qgx.png)
Multiplying the first and the second parentheses:
![f(x)=(x^4+(4+9)x^2+(4)(9))(x-1)\\ f(x)=(x^4+13x^2+36)(x-1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/q65ivcm7m6ovo80320t9dvnvje07jwrund.png)
Multiplying the parentheses:
![f(x)=x(x^4)+x(13x^2)+x(36)-1(x^4)-1(13x^2)-1(36)\\ f(x)=x^5+13x^3+36x-x^4-13x^2-36](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o2njtgo4smo6u8vzt6dq8d9ppr5u0iv4s5.png)
Ordering the terms:
![f(x)=x^5-x^4+13x^3-13x^2+36x-36](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9co5umi0plkf6qdimgzbwqfpyaaxulwfar.png)