Reducible and Irreducible polynomials over $\Bbb Q$ Consider polynomials of the form: $$x^r-(1-x)^k,$$
for $r,k\ge2.$
$x\in(0,1).$
When $r=k$ the polynomial seems to be reducible, except at $r=k=2.$ 
Do the irreducible and reducible polynomials form a pattern when plotted?  
To try to visualise what was going on, I made a lattice of all the points representing reducible and irreducible polynomials of this form:
$$ x^r=(1-x)^k. $$
I plotted each solution $x$ at a certain height $h$ in such a way that the solutions formed a grid.
For example, all points greater than $2$ for $r=k,$ I colored green, because they are reducible, and plotted an $x$ value of $1/2$ and at different heights $h.$
Here's what it looks like for $r,k=\{2,3,4,5,6\}.$ Red=Irreducible, Green=Reducible.



Do the irreducible and reducible polynomials form a pattern when plotted?

Depends what you consider a pattern, but here are few quick observations. Let $f_{r,k}(x)=x^r-(1-x)^k$, then:


*

*The graph is vertically symmetrical. This follows simply from fact that if $f(x)$ is irreducible, then $f(-x)$ is irreducible, as well as $f(x+1)$. In this case both together can be used to see $f(1-x)$ is irreducible iff $f(x)$ is irreducible, which applied to $f_{r,k}(x)=f_{k,r}(-x+1)$ gives you the vertical symmetry.

*Middle vertical line is all consisting of green dots except for $(r,k)=2$. This is because for $r=k$ we have $f_{r,r}(x)=x^r-(1-x)^r$ is always multiple of $(2x-1)$  hence reducible. To see this (I'm sure there are easier ways), consider for example
\begin{align*}
f_{r,r}(x)&=x^r-(1-x)^r\\
&=(2x-1+(1-x))^r-(1-x)^r\\
&=\sum \binom{r}{i}(2x-1)^i (1-x)^{r-i}-(1-x)^r.
\end{align*}
All of the terms in the sum are multiples of $2x-1$ except for one with $i=0$, and so for some polynomial $P(x)$ we have
\begin{align*}
f_{r,r}(x)&=P(x)(2x-1)+(1-x)^r-(1-x)^r=P(x)(2x-1).
\end{align*}

*There is a vertical line going through just red dots. There is a family of polynomials $f_{r+1,r}(x)$ which seem to be irreducible (as well as they symmetrical counterpart $f_{r,r+1}(x)$). I could not find a proof suitable for all $r$'s, but I've checked for $r\leq 1000$ and they are all irreducible. Maybe someone will find a proof, but notice that such sequences of irreducible polynomials are quite common, the real challenge is to prove it. For example consider:


*

*Is $ f_n=\frac{(x+1)^n-(x^n+1)}{x}$ irreducible over $\mathbf{Z}$ for arbitrary $n$?

*Is polynomial $x^{2n-1}-4nx^{2n-2}-2\binom{2n}{3}x^{2n-4}-\dots-2\binom{2n}{3}x^{2}-4n$ irreducible? 

*Irreducibility of q-factorial plus 1

*there are many more...


