What is the polynomial function of degree 4 with zeros of 3, -1, and 2-i where f(0)=30?

Answer 1

The problem is unsolvable in its current formulation. However, if the quartic is to have real coefficients, then it is solvable. In that case, since one root is complex, another root is its complex conjugate: 2 + i.

For some constant a, P(x) must look like

P = a (x - 3) (x +1) (x-2+i) ( x-2-i) = a (x - 3) (x+1) (x² + 5)

For P(0) to be 30, we have

30 = a (-3)(1)(5)

a = -2

Thus P(x) = -2 (x -3) (x+1) (x²+5), ans.