Question 840326
To solve this for x, you first need to perform the rational zero test, to determine all of the possible real zeros.  Doing this will give you the following possible zeros:


+-1, +-2, +-3, +-6, +-1/3, +-2/3


Next you will have to test these possible zeroes by synthetic division.  It's easier to start with the small integers first.  If you start with 1, you will see that 1 IS a zero of this polynomial.  Your remaining polynomial when you synthetically divide by 1, is


{{{3x^2+9x+6}}}


The first thing you can do is to factor out the 3 from that remaining polynomial, which will give you


{{{3(x^2+3x+2)}}}


Now, you can easily factor.  Doing so, will give you


{{{3(x+2)(x+1)}}}


Set each of these factors equal to zero.  You can remove the 3, because 3 = 0 is not a correct statement.  Doing this will give you


x + 2 = 0  and x + 1 = 0


Solving for x in both of the above, will give you your other two zeros:


x = -2 and x = -1


Thus, the zeros for {{{3x^3+6x^2-3x-6}}} are


-2, -1, 1