Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 330: 40

$P(x)=3x^{3}+3x$

Since $0, i$ are zeros, then so $-i$ by the Conjugate Zeros Theorem. This mean that P(x) have the following form: $P(x)=ax[x-i][x-(-i)]$ $P(x)=ax(x-i)(x+i)$ $P(x)=a(x^{2}-xi)(x+i)$ $P(x)=a(x^{3}+x^{2}i-x^{2}i-xi^{2})$ $P(x)=a(x^{3}-xi^{2})$ $P(x)=a(x^{3}+x)$ To make all all coefficients interger, we set $a=3$ and get $P(x)=3x^{3}+3x$