Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?

Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations.

 A: $$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$
$$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$
$$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$
$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$
A: We have $\displaystyle x+10x^3+100x^5=x\frac{1000x^6-1}{10x^2-1}$. (A geometric progression)
Hence $\displaystyle -2(1000x^6-1)=x \frac{1000x^6-1}{10x^2-1}$
Hence either $1000x^6-1=0$ or $x=-2(10x^2-1)$.
Therefore $20x^2+x-2=0$ for second equation.
Solving we get 
$$x=\frac{-1\pm \sqrt{161}}{40}$$.
Comparing $m=-1, n=161$ and $r=40$. Hence $m+n+r=200$
A: HINT: try the ansatz $$(-2+Bx+Ax^2)(1+Cx^2+Dx^4)$$
A: The usual trick is to divide through by $x^3.$ Then notice that taking $w = 10x - \frac{1}{x}$ seems to allow writing the thing, and we get
$$ 2 w^3 + w^2 + 60 w + 30 = 0.  $$
This has a rational root, namely $-1/2,$ and factors as
$$ (2w+1)(w^2 + 30)  $$
Setting
$$ 10 x - \frac{1}{x} = -\frac{1}{2}  $$
leads to
$$ 20 x^2 + x - 2 = 0 $$
