Polynomial Equations Root Finding If one of the roots of the Equation : 
$$2000x^6 + 100x^5 + 10x^3 + x - 2 = 0$$ ;
Is of the form $$(m + √n)/r$$. Where m is a non zero integer and n and r are relatively coprime. 
We have to find $$(m+n+r)/100$$ .
I am not able to factorise this as I am not able to guess any roots.  
I could only comment that there's a root between 0 and 1 which doesn't help
Using sum , products of roots seems too tough to even start...
Any hint would do ... 
Thanks in advance.
 A: Given $$2000x^6+100x^5+10x^3+x-2=0\Rightarrow (2000x^6-2)+(100x^5+10x^3+x)=0$$
Now Divide it by $x^3\;,$ Where $x\neq 0\;,$ We get
$$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\left[(10x)^2+1+\frac{1}{x^2}\right]=0$$
So $$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\frac{(10x)^3-\frac{1}{x^3}}{\left[10x-\frac{1}{x}\right]}=0$$
So $$\left[(10x)^3-\frac{1}{x^3}\right]\cdot \left[2+\frac{1}{\left(10x-\frac{1}{x}\right)}\right]=0$$
So $$\left[(10x)^2+1+\frac{1}{x^2}\right]\cdot \left(20x^2+x-2\right)=0$$
So we get $100x^4+10x^2+1=0$ or $20x^2+x-2=0$
Now $100x^4+10x^2+1>0\;\forall x\in \mathbb{R}$. So here $$\displaystyle 20x^2+x-2=0\Rightarrow x = \frac{-1\pm \sqrt{1+160}}{2\times 20} = \left(\frac{-1\pm \sqrt{161}}{40}\right)$$
So after camparing with $\displaystyle \frac{m+\sqrt{n}}{r}\;,$ We get $m+n+r=-1+61+40=200$
So $$\displaystyle \frac{m+n+r}{100} = \frac{200}{2} =2$$
A: By Descartes rule of signs, we have that the equation has exactly $1$ positive root,exactly $1$ positive root and $4$ complex roots.
As the coefficients are all real and rational, the $4$ complex roots are complex conjugates and the $2$ real roots are conjugate surds as one of the real roots has been given to be a quadratic surd.
Hence the other real root is $\frac{m-\sqrt n}{r}$.
Does this help if you now use the relation between coefficients of terms and the roots?
