Let’s solve the even-numbered problems from 6 to 16 as requested.

6. Zeros: $1 - 2i$ and $1 + 2i$

Since the zeros are complex conjugates and the coefficients are real: $f(x) = (x - (1 - 2i))(x - (1 + 2i))$ Expand: $f(x) = (x - 1 + 2i)(x - 1 - 2i) = (x - 1)^2 - (2i)^2 = (x - 1)^2 + 4$ Simplify: $f(x) = x^2 - 2x + 1 + 4 = x^2 - 2x + 5$

8. Zeros: $-4$, $1 - i$, and $1 + i$

With $1 - i$ and $1 + i$ being conjugates, the polynomial is: $f(x) = (x + 4)(x - (1 - i))(x - (1 + i))$ Expand: $(x - (1 - i))(x - (1 + i)) = (x - 1 + i)(x - 1 - i) = (x - 1)^2 - i^2 = (x - 1)^2 + 1$ $f(x) = (x + 4)((x - 1)^2 + 1)$ Simplify: $f(x) = (x + 4)(x^2 - 2x + 1 + 1) = (x + 4)(x^2 - 2x + 2)$ Distribute: $f(x) = x^3 - 2x^2 + 2x + 4x^2 - 8x + 8 = x^3 + 2x^2 - 6x + 8$

10. Zeros: $-1$, $2$, and $1 - i$, $1 + i$

Include conjugates $1 - i$ and $1 + i$: $f(x) = (x + 1)(x - 2)(x - (1 - i))(x - (1 + i))$ Expand $x - (1 - i)$ and $x - (1 + i)$: $(x - (1 - i))(x - (1 + i)) = (x - 1)^2 + 1 = x^2 - 2x + 2$ $f(x) = (x + 1)(x - 2)(x^2 - 2x + 2)$ Simplify: $f(x) = (x^2 - x - 2)(x^2 - 2x + 2)$ Distribute: $f(x) = x^4 - 2x^3 + 2x^2 - x^3 + 2x^2 - 2x - 2x^2 + 4x - 4$ $f(x) = x^4 - 3x^3 + 2x^2 + 2x - 4$

12. Zeros: $-2$, $1 + 2i$, and $1 - 2i$

With $1 + 2i$ and $1 - 2i$ as conjugates: $f(x) = (x + 2)(x - (1 + 2i))(x - (1 - 2i))$ Expand $x - (1 + 2i)$ and $x - (1 - 2i)$: [ (x - (1 + 2i))(x - (1 -