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?
 A: 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 $p(y)$. 
A: 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 
$$[(x+1)^2-3^2]+(x+2)^3+2^3+(x+3)^4-1^4$$
$$=(x+4)(x-2)+\{(x+2)+2\}[(x+2)^2-2(x+2)+2^2]+[(x+4)(x+2)\{(x+3)^2+1\}]$$
A: 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 $x=-2$.
A: Set $y=x+2$. The equation rewrites as:
$$(y-1)^2+y^3+(y+1)^3==2\iff y(y^3+5y^2+7y^+2)=0$$
The second factor has an integer root: $-2$, hence the factorisation:
$$y^3+5y^2+7y +2=(y+2)(y^2+3y+1).$$
Finally, the roots (in $y$) are: $0, -2, \dfrac{-3\pm\sqrt 5}{2} $, whence the roots of the initial equation:
$$ \Bigl\{ -2,-4, \frac{-7\pm\sqrt 5}{2}\Bigr\}.$$
