Question

Find an nth degree polynomial function with real coeffic n=3;3 and i are zeros; f(2)=25

Answer 1

Hi,

Explanation:

3 and i are zeros.

Since all coef are real: -i is also a zero.

`P(x)=k*(x-3)(x-i)(x+i)=k(x-3)(x^2+1\\P(2)=25=k*(2-3)(2^2+1)\\k=-5\\\\\boxed{P(x)=-5(x-3)(x^2+1)}\\`