<s>[INST] 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?
 [/INST] Hint: Write
$$(x + 3 - 2)^2 + (x+3 - 1)^3 + (x+3)^4 = 2$$
Let $y = x+3$ then 
$$p(y) = y^2 - 4y + 4 + y^3 - 3y^2 + 3y -1 + y^4 = 2 \implies p(y) =y^4 + y^3 -2y^2 - y +1 = 0 $$
Notice that $1$ and $-1$ are  roots of $p(y)$. 
</s>