Question 1003820
<pre>
{{{x^4+x^3-5x^2-3x+6=0}}}

The potential rational roots are ± the factors of the
last term (in absolute value).

±1, ±2, ±3, ±6

We try 1

1 |1  1 -5 -3  6
  |<u>   1  2 -3 -6</u>
   1  2 -3 -6  0

So we have factored the equation as:

{{{(x-1)(x^3+2x^2-3x-6)=0}}}

We can factor the expression in the second parentheses
by grouping, ie., factor the first two terms and factor
the last two terms:

{{{(x-1)(x^2(x+2)-3(x+2))=0}}}

Factor (x+2) out of the big parentheses:

{{{(x-1)(x+2)(x^2-3)=0}}}

Use zero-factor property:

{{{x-1=0}}}; {{{x+2=0}}}; {{{x^2-3=0}}}
{{{x=1}}};    {{{x=-2}}};  {{{x^2=3}}}
                   {{{x="" +- sqrt(3)}}}

The irrational roots are {{{sqrt(3)}}} and {{{-sqrt(3)}}}.

Their sum is 0.

Edwin</pre>