Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 294: 47

$2i,-2i,-2$

Consider $P(x)=x^3+2x^2+4x+8=0$ In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows: $x^3+2x^2+4x+8=x(x^2+4)+2(x^2+4)$ This gives: $(x+2)(x^2+4)=0$ This can be re-written as: $(x^2-4i^2)(x+2)=0 \implies (x^2-(2i)^2)(x+2)=0$ $(x-2i)(x+2i)(x+2)=0$ Thus the zeroes of $P(x)$ are: $2i,-2i,-2$