Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 330: 47

$x\in \{-2, -2i, 2i\}$

$P(x)=x^3+2x^2+4x+8$. Factor the polynomial by grouping in pairs: $$\begin{aligned} P(x)&=x^3+2x^2+4x+8\\ &=x^2(x+2)+4(x+2)\\ &=(x^2+4)(x+2). \end{aligned}$$ Solve the equation $P(x)=0$: $(x^2+4)(x+2)=0$, either $(x^2+4)=0, x=\pm2i$ or $x+2=0, x=-2$. thus the zeros are $$x\in \{-2, -2i, 2i\}$$