Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 294: 42

$x^3+x^2-4x+6$

The Complex Conjugate zeroes Theorem states that the conjugate of $i$ is also a zero of the polynomial $P(x)$. Here, we have three zeroes of the polynomial $P(x)$ of degree $3$. The factorization of $P(x)$ is given as follows: $P(x)=(x+3)[x-(1+i)][x-(1-i)]$ Apply the difference square formula. $P(x)=(x+3)[x-(1+i)][x-(1-i)]=(x+3)(x^2-2x+1+1)=(x+3)(x^2-2x+2)=x^3+x^2-4x+6$