<PRE><B>Question 364377</B>
&lt;pre&gt;&lt;b&gt;
When you are told that that a surd a + b&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;c is a solution, 
you also know that its conjugate a - b&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;c is also a soution. 
 
Since 2 - &lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5 is a zero, we&#39;ll synthetically divide

x³ - x² - 13x - 3 by 2 - &lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5


2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5|1   -1    -13     -3
    |&lt;u&gt;     2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5   7-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5  3&lt;/u&gt;
     1    1-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5  -6-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5  0  
 
You&#39;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-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5), so you multiply
that by the number at the far left, 2-V5 by FOIL:

         (1-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)(2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5) = 2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5-2&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5+5 = 7-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5
 
and the others are similar.

So now the left side of the equation has been factored as:

[x - (2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)][x² + (1-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)x + (-6-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)] = 0

The conjugate of x - (2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5) is x + (2+&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5), and it is
a solution too, so since 2 + &lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5 is also a zero, we&#39;ll 
synthetically divide

x² + (1-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)x + (-6-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)  by 2+&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5

2+&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5|1     1-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5    (-6-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5)
    |______________________


Remove the parentheses and do the synthetic division:

2+&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5|1    1-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5   -6-3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5
    |&lt;u&gt;&lt;b&gt;     2+&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5    6+3&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5&lt;/u&gt;                 
     1    3       0
     
So we have completely factored the left side of the equation: 

[x - (2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5][x + (2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5](x + 3) = 0

The three zeros of the function difined by the left side
(solutions to the equation), are 

 2-&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5, 2+&lt;font face=&quot;symbol&quot;&gt;Ö&lt;/font&gt;5, and -3.

Edwin&lt;/pre&gt; 

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