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?

Although the substitution $y=x+2$ works well, I would first try $y=x+3$, which avoids having to expand the highest power (quartic) term and leads to $$y^4+y^3-2y^2-y+1=0$$This yields two roots $y=\pm 1$ to the rational root theorem. Or alternatively to the observation that $$y^4+y^3-2y^2-y+1=(y^4+y^3-y^2)-(y^2+y-1)=(y^2-1)(y^2+y-1)$$
A second thought would be to spot the even powers $(x+1)^2, (x+3)^4$, and to see that the minimum value of the sum of these for integer $x$ is $2$, which would lead to spotting the root $x=-2$.
