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

Question

n = 3

3 and 4 i are zeroes

f(1) = -102

f(x) =

Answer 1

The complex conjugate of a complex zero is also a zero, so your 3 zeros are

3

4i

-4i

f(x) = A(x - 3)(x - 4i)(x + 4i)

f(x) = A(x - 3)(x² + 16)

f(1) = -102 ⇒

-102 = A(1 - 3)(1² + 16)

Now solve for A and finish from here.