How to find a second complex root when factoring a polynomial using synthetic division

A few of the problems involved synthetic division.

They might start as third degree or fourth degree polynomials with one or more of the roots given. There was an equation that would give a second root if one was a complex number (if all the coefficients of the original polynomial are real).

The complex conjugate will also be a root.

For example, if 1 + i is a root, then 1- i is also a root. (given the right conditions)

If you have two roots and use synthetic division (or long division) you will have two roots and a quadratic equation, which will be much easier to factor.