Roots of equation $x^n\left(2-x\right)^{2}=a$? I need to find analytically the roots of the following polynomial equation:
$$x^n\left(2-x\right)^{2}=a$$
for an arbitrary integer $n$ and an arbitrary real parameter $a$. The only trick I can think of is in the special case when $a\geq0$ and $n=2m$ (i.e. $n$ even). In this case the equation can be decomposed into two simpler and lower dimensional equations:
$$x^{m}\left(2-x\right)-\sqrt{a}=0$$
$$x^{m}\left(2-x\right)+\sqrt{a}=0$$
but then I don't know how to proceed (I was thinking that maybe through some change of coordinates it is possible to transform these new equations into trinomial equations of the form $x^n-x+t=0$, whose solution is known, see here). I'd also like to find the roots for any $n$ and $a$. Closed-form solutions are not required: I'd really appreciate also a solution written e.g. as a series expansion.
Many thanks in advance.
P.S.: I don't know if it may help, but in my case $x\in[-1,1]$ so that you may write $x=\cos y$ for $y\in [0,\pi]$.
 A: The first thing you can do is to study the trivial case $a=0$ which gives :
$$x^n(2-x)^2=0$$
The roots are obviously $x=0$ with order $n$ and $x=2$ with order $2$. You can try to solve other cases to sketch the general form of the solution if it exists.
You can also try some particular values of $n$ :
For $n=0$ you have :
$$x=2\pm\sqrt{a}$$
Which is an order $2$ root.
For $n=1$ I tried with Wolfram Alpha which provides only one ugly solution, which has to be order $3$
For $n=2$ you have : 
$$x^2(2-x)^2=a$$
$$x(2-x)=2x-x^2=\pm\sqrt{a}$$
Hence the solutions are :
$$x=1\pm\sqrt{1\pm\sqrt{a}}$$
Depending on $a$ you can have up to four different solutions with order $1$ each.
It seems complicated to find a general expression as you can have different number/orders of solutions depending on the choice of $(a,n)$ 
A: See the following link (theorem 2) https://www.researchgate.net/publication/262973394_Solution_of_Polynomial_Equations_with_Nested_Radicals
He solves :
$$Aqx^{p}+x^{q}=1$$
Your equation is :
$$2x^m-x^{m+1}-\sqrt{a}=0$$
If you make the following substitution :
$x=y\beta$
We have :
$$2\beta^{m}y^m-\beta^{m+1}y^{m+1}-\sqrt{a}=0$$
So divide by $-\beta^{m+1}$ you have:
$$\frac{-2}{\beta}y^m+y^{m+1}+\frac{\sqrt{a}}{\beta^{m+1}}=0$$ 
And make the last substitutions :
$$\beta^{1}=\frac{1}{m+1}$$
and
$$\sqrt{a}=\beta^{m+1}$$
Finally we obtain :
$$(-2)(m+1)y^m+y^{m+1}+1=0$$ 
So you can apply the theorem 2 with $A=-2$   and   $q=m+1$
Another way is to use the theorem 1 with $b=0$ but we obtain a nested radical...
Ps:
It's a partial solution because there is some conditions on $\sqrt{a}$...
If you want other details see this Solving 5th degree or higher equations
A: Excel can provide some intuition. Here are graphs of $x^n(2-x)^n$ on $[-1,1]$ for $n=6$ and $n=7$. 

You can see the range in each case, and prove it (for $n$ odd and $n$ even). For large(r) $n$ than these modest values the shape remains the same, so for values of $a$ (not too close to $0$) you'll have one or two solutions. Those solutions approach $\pm 1$ as $n$ grows. Newton's method should work just fine.
