<PRE><B>Question 1010528</B>
x^4+x^3-6x^2-14x-12=0.  Roots possible include +/- 12,6,4,3,and 2.  It has one sign change, so it has one positive root.  f(-x) +/-/-/+/-, so there are 3 or 1 negative roots.
1 ;;1;;-6;;-14;;-12
1;;-1;;-4;;-6  synthetic division using -2 as a root or (x+2)
now we have x^3-x^2-4x-6
1;;-1;;-4;;-6  try 3
1;;2;;2;;0.  That works (x-3)
x^2+2x+2. That has complex roots.

{{{graph(300,200,-10,10,-10,100,x^4+x^3-6x^2-14x-12)}}}

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