Question 319327
<pre><b>
{{{"g(x)"=x^3-3x^2-3x+1}}}

Move the "+1" term:

{{{"g(x)"=x^3+1-3x^2-3x}}}

Factor the first two terms, {{{x^3+1}}}, as {{{(x+1)(x^2-x+1)}}}
Factor the last two terms, {{{-3x^2-3x}}}, as {{{-3x(x+1)}}}


{{{"g(x)"=(x+1)(x^2-x+1)-3x(x+1)}}}

Factor out {{{(x+1)}}}

{{{"g(x)"=(x+1)}}}{{{"["}}}{{{(x^2-x+1)-3x}}}{{{"]"}}}

{{{"g(x)"=(x+1)(x^2-x+1-3x)}}}

{{{"g(x)"=(x+1)(x^2-4x+1)}}}

Now we have to find the two zeros of {{{x^2-4x+1}}}
which is not factorable with integers.  So we use the 
quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-4) +- sqrt( (-4)^2-4*(1)*(1) ))/(2*(1)) }}}
{{{x = (4 +- sqrt(16-4))/2 }}}
{{{x = (4 +- sqrt(12))/2 }}} 
{{{x = (4 +- sqrt(4*3))/2 }}}
{{{x = (4 +- 2*sqrt(3))/2 }}}
{{{x = 4/2 +- 2*sqrt(3)/2 }}}
{{{x = 2 +- sqrt(3) }}}

So the other two zeros are 

{{{x = 2 + sqrt(3) }}} and {{{x = 2 - sqrt(3) }}}

or

{{{x - 2 - sqrt(3)=0 }}} and {{{x - 2 + sqrt(3) }}}

So now 

{{{"g(x)"=(x+1)(x^2-4x+1)}}}

becomes:

{{{"g(x)"=(x+1)(x-2-sqrt(3))(x-2+sqrt(3))}}}

Edwin</pre>