Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?

HINT:
$$[(x+1)^2-1]+(x+2)^3+(x+3)^4-1^4$$
$$=x(x+2)+(x+2)^3+[\{(x+3)^2-1\}\{(x+3)^2+1\}]$$
$$=x(x+2)+(x+2)^3+[(x+4)(x+2)\{(x+3)^2+1\}]$$
and 
$$[(x+1)^2-3^2]+(x+2)^3+2^3+(x+3)^4-1^4$$
$$=(x+4)(x-2)+\{(x+2)+2\}[(x+2)^2-2(x+2)+2^2]+[(x+4)(x+2)\{(x+3)^2+1\}]$$
