All possible rational solutions are ± divisors of 4
These are
±1, ±2, ±4
We begin by trying 1. To use synthetic division we must
consider the equation to be:
1 | 1 0 -6 4
| 1 1 -5
1 1 -5 -1
Nope. The remainder is -1 not 0, so 1 is not a solution.
We try -1:
-1 | 1 0 -6 4
| -1 1 5
1 -1 -5 9
Nope. The remainder is 9 not 0, so -1 is not a solution.
We try 2:
2 | 1 0 -6 4
| 2 4 -4
1 2 -2 0
Yep! The remainder is 0, so x=2 is a solution. Therefore
we have factored the left side as
Using the zero-factor principle:
Setting 
of course gives the solution we just found,
namely 
Setting
This does not factor, so we use the quadratic formula
with 
, 
, 
, 
, and 
That's 3 real solutions. No imaginary solutions.
Edwin
