Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 330: 42

$Q(x)=x^3+x^2-4x+6$

The factor theorem says that if $f(c)=0$, then $(x-c)$ is a factor of $f(x)$ and if $(x-c)$ is a factor of $f(x)$, then $f(c)=0$. According to the Conjugate Pair Theorem, since $1+i$ is a complex zero, $1-i$ is also a complex zero. We use the zeros to construct factors, which we multiply to find the original equation: $Q=(x-(-3))(x-(1+i))(x-(1-i))=x^3+x^2-4x+6$