Factor $x^5-x+15$ It's possible to factor $x^5-x+15$. WolframAlpha gives the answer of: $$(x^2+x+3)(x^3-x^2-2x+5)$$
According to the wikipedia article on quintic functions, the general form $x^5-x+a$ is factorable only when $a=±15$, $±22440$, or $±2759640$.
Question: How would one factor such an expression? For me, it seems close to impossible by hand.
Bonus question: Why do those specific values of $a$ make the expression factorable? If there's no simple answer, is there a paper/further reading about it?

I see there is an answer to 15. I tried, it appears there is no integer root to $x^5 - x \pm 22440.$ As 7 gives 16800 but 8 gives 32760. We arrive at
$$ (x^3 + A x^2 + B x + C)(x^2 + D x + E) = x^5 - x \pm 22440 $$
The point is not to solve the whole system at once, rather do one coefficient at a time and rewrite the system.  The degree four term must be 0, so $A+D = 0$
$$ (x^3 + A x^2 + B x + C)(x^2 -A x + E) = x^5 - x \pm 22440 $$
Next the cubed term is zero, so $E-A^2 + B = 0, $ or $E = A^2 - B.$
$$ (x^3 + A x^2 + B x + C)(x^2 -A x + (A^2 - B)) = x^5 - x \pm 22440 $$
Next $x^2$ has 0, or $A^3 - AB -AB + C = 0,$ or $C = 2AB - A^3,$
$$ (x^3 + A x^2 + B x + (2AB -A^3))(x^2 -A x + (A^2 - B)) = x^5 - x \pm 22440 $$
Linear coefficient is $-1,$ so $A^2 B - B^2 - 2 A^2 B  + A^4 = -1,$ or $A^4 - A^2 B - B^2 = -1.$ 
Getting there; taking $x = A^2, y = B,$ we have $x^2 - xy - y^2 = -1,$ meaning that $x,y$ are consecutive Fibonacci numbers... as $(3,2), (8,5), (21,13), (55,34), (144, 89).$ Since $x$ needs to be a square, we will try $A^2 = 144,$ $A = 12,$ $B = \pm 89$
One selection is
$$ (x^3 + 12 x^2 + 89 x + 408)(x^2 - 12 x + 55)  = x^5 - x + 22440$$
Just negating $x$ gives
$$ (-x^3 + 12 x^2 - 89 x + 408)(x^2 + 12 x + 55)  = -x^5 + x + 22440,$$
$$ (x^3- 12 x^2 + 89 x - 408)(x^2 + 12 x + 55)  = x^5 - x - 22440,$$
So there you go. We were able to use the first Fibonacci number that is also a square. Bigger than $1$ I guess. 
