Question 309868
{{{x^3 - 6x +4 = 0}}}
<pre><font size = 4 color = "indigo"><b>
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:

{{{x^3 +0x^2- 6x + 4 = 0}}}

1 | 1  0 -6  4
  |<u>    1  1 -5</u>
    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
   |<u>   -1  1  5</u>
     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
   |<u>    2  4 -4</u>
     1  2 -2  0

Yep! The remainder is 0, so x=2 is a solution.  Therefore
we have factored the left side as

{{{(x-2)(x^2+2x-2)=0}}}

Using the zero-factor principle:

Setting {{{x-2=0}}} of course gives the solution we just found,
namely {{{x=2}}}

Setting 

{{{x^2+2x-2=0}}}

This does not factor, so we use the quadratic formula
with {{{a=1}}}, {{{b=2}}}, {{{c=-2}}}

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

{{{x = (-(2) +- sqrt( (2)^2-4*(1)*(-2) ))/(2*(1)) }}}

{{{x = (-2 +- sqrt(4+8))/2 }}}

{{{x = (-2 +- sqrt(12))/2 }}}

{{{x = (-2 +- sqrt(4*3))/2 }}}

{{{x = (-2 +- sqrt(4)sqrt(3))/2 }}}

{{{x = (-2 +- 2*sqrt(3))/2 }}}

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

{{{x = (cross(2)(-1 +- sqrt(3)))/cross(2)}}}

{{{x = -1 +- sqrt(3) )}}

So the solutions are {{{2}}}, {{{-1+sqrt(3)}}}, and {{{-1-sqrt(3)}}}

That's 3 real solutions.  No imaginary solutions.

Edwin</pre>