Chapter 3 - Polynomial and Rational Functions - Section 3.3 Complex Zeros; Fundamental Theorem of Algebra - 3.3 Assess Your Understanding - Page 231: 28

$1-3i,-1,6$.

Step 1. As $x=1+3i$ is a zero, we have $x=1-3i$ also a zero and $(x-1-3i)(x-1+3i)=(x-1)^2+9=x^2-2x+10$ is a factor of the function. Step 2. Use synthetic division (two steps) or long division (as shown in the figure) to get the quotient. Step 3. Find other zeros $x^2-5x-6=0 \Longrightarrow (x-6)(x+1)=0 \Longrightarrow x=-1,6$.