Question 1056898
General form of a 4th degree polynomial is:

    {{{ax^4 + bx^3 + cx^2 + dx + e  = 0 }}}

This can be constructed from the zeros:
 
The zero at -3 means there is a factor (x+3) (because plugging in x=-3 gives zero)
The zeros at 2,-2, and -1 yield three more factors:  (x-2)(x+2) and (x+1), respectively.

Putting all the factors together:
                (x+3)(x-2)(x+2)(x+1)
            = (x+3)(x+1)(x+2)(x-2)     (reordered, will multiply first two, last two next)
             = {{{(x^2+4x+3)*(x^2-4)}}}
             = {{{x^4+4x^3+3x^2-4x^2-16x-12}}}
             = {{{x^4+4x^3-x^2-16x-12}}}

 The problem said the coefficient of {{{x^2}}} should be -3, since it is already -1, we need to multiply the polynomial by 3:

              Ans: {{{3x^4+12x^3-3x^2-48x-36}}}