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

Question

N=3;

2 and 5 I are zeros;

f (1)=-52

Answer 1

then -5i is also a solution

So P(x) = C(x-2)(x+5i)(x-5i)

= C(x-2)(x^2 + 25)

(1, -52) ---> -52 = C ( -1)(26) ---> C = 2

P(x) = 2(x-2)(x^2 + 25) = 2( x^3 -2x^2 + 25x - 50)

= 2x^3 - 4x^2 + 50x - 100