degree 5; zeros: -4;-i;4+i

Question

formula of polynomial

Answer 1

Only three zeroes are provided for the degree 5 polynomial. Two zeroes are complex, so for the polynomial to be real, the remaining two zeroes, I assume, must counterbalance the two complex ones. Therefore the two remaining zeroes are i and 4-i. The factors are:

(x+i)(x-i)(x-4+i)(x-4-i)(x+4)=(x^2+1)(x^2-8x+17)(x+4)⇒(x^4-8x^3+18x^2-8x+17)(x+4)

⇒x^5-8x^4+18x^3-8x^2+17x+4x^4-32x^3+72x^2-32x+68⇒x^5-4x^4-14x^3+64x^2-15x+68.