Solve $2000x^6+100x^5+10x^3+x-2=0$

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 $m+n+r=200$

HINT: try the ansatz $$(-2+Bx+Ax^2)(1+Cx^2+Dx^4)$$

$$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$ $$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$ $$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$

$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$