SOLUTION: Use synthetic division to show that x is a solution of the third- degree polymonial equation, and use the result to factor the polymonial completely. List all real zeros of the

When you are told that that a surd a + bÖc is a solution, 
you also know that its conjugate a - bÖc is also a soution. 
 
Since 2 - Ö5 is a zero, we'll synthetically divide

x³ - x² - 13x - 3 by 2 - Ö5


2-Ö5|1   -1    -13     -3
    |     2-Ö5   7-3Ö5  3
     1    1-Ö5  -6-3Ö5  0  
 
You've probably done synthetic division only when
there were just integers and no square roots

The multiplications required to do this is just the same way 
except every time you have to get some scratch paper. When you multipy 
each number on the bottom times the divisor number at the far left, you 
have to use FOIL on the side of your paper or on scratch paper. You have
to do this every time you multiply after the first step when you multiply 
by 1.  Then you add the second column and get (1-Ö5), so you multiply
that by the number at the far left, 2-V5 by FOIL:

         (1-Ö5)(2-Ö5) = 2-Ö5-2Ö5+5 = 7-3Ö5
 
and the others are similar.

So now the left side of the equation has been factored as:

[x - (2-Ö5)][x² + (1-Ö5)x + (-6-3Ö5)] = 0

The conjugate of x - (2-Ö5) is x + (2+Ö5), and it is
a solution too, so since 2 + Ö5 is also a zero, we'll 
synthetically divide

x² + (1-Ö5)x + (-6-3Ö5)  by 2+Ö5

2+Ö5|1     1-Ö5    (-6-3Ö5)
    |______________________


Remove the parentheses and do the synthetic division:

2+Ö5|1    1-Ö5   -6-3Ö5
    |     2+Ö5    6+3Ö5                 
     1    3       0
     
So we have completely factored the left side of the equation: 

[x - (2-Ö5][x + (2-Ö5](x + 3) = 0

The three zeros of the function difined by the left side
(solutions to the equation), are 

 2-Ö5, 2+Ö5, and -3.

Edwin