Section 2.5 Question 44 [ATTACH=full]470[/ATTACH] We are given 4 zeros. This leads to a fifth degree equation. Why is this the case? In terms of 2i, the complex conjugate must also be a zero. We now have this: f(x) = a(x +1)(x - 5)(x - 3)(x - 2i)(x + 2i) f(x) = a(x^3 - 7x^2 + 7x + 15)(x^2 + 4) I know that a = 1 but what is the reason for this value of a? Why does a = 1? f(x) = x^5 -7x^4 + 11x^3 - 13x^2 + 28x + 60 You say?