write a polynomial function f of least degree that has rational coefficient of 1, and the given zeros 3i, 2, -i it should be written like f(x)=....

Answer 1

change the sign of each zero, to -3i, -2 and i. Then add x to each. Then multiply them all together

f(x) = (x-3i)(x-2)(x+i). That has lead coefficient of 1 for the x^3 term, when expanded, which is the least degree possible.

But usually the imaginary roots show up as conjugate pairs, so if 3i is a root or zero, then so is -3i. if so,

then f(x) = (x^2+9)(x-2)(x^2+1) which also has lead coefficient of 1 for x^5 term. That's the least degree polynomial that has real coefficients.

(x-3i)(x+3i) = x^2+9, eliminating the imaginary coefficients you'd have in a 3 degree polynomial.

same with (x-i)(x+i) = x+1