find an nth degree polynomial function with real coefficients satisfying the given conditions n=3

Question

-4 and 4+5i are zeros

f(1)= 170

Answer 1

Since n = 3 and we are given two zeroes:

(x+4)(x-4-5i)(x+a+ib)

f(1) = (5)(-3-5i)(1+a+ib) =170 --> (3+5i)(1+a+ib) = -34

Expanding

3 + 3a + 3bi +5i + 5ai -5b = -34

Separating Reals and Imaginaries

3a -5b = -37

5a +3b = -5

Solving the two simultaneous equations gives

a = -4 and b = 5

(x+4)(x-4-5i)(x-4+5i) = (x+4)(x²-4x+5ix-4x+16 -20i-5ix+20i +25) = (x+4)(x²-8x +41)

= x³-8x²+41x +4x²-32x+164 = x³-4x²+9x+164

Answer 2

(x+4)(x^2-8x+41) =x^3 -4x^2 + 9x + 161

f(1) = 1 -4 + 9+ 164 = 170