Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.

Question

n=3

2 and 4i are zeroes

f(-1)=-51

Answer 1

Since the degree is 3, there are at most 3 distinct roots.

Since the polynomial has real coefficients and 4i is a root, then so is -4i.

f(x) = a(x - 2)(x - 4i)(x - (-4i)) = a(x - 2)(x - 4i)(x+ 4i)

= a(x - 2)(x² + 16)

f(-1) = -51, so a(-3)(17) = -51. Therefore, a = 1

f(x) = (x - 2)(x² + 16)

f(x) = x³ - 2x² + 16x - 32