When All Else Fails, We Don't

Let $f(x) = 2000x^6 + 100x^5 + 10x^3 + \color{#3D99F6}{x - 2}$ and assume that it can be factorised as follows.

$\begin{aligned} f(x) & = (ax^4 + bx^2+\color{#3D99F6}{1})(cx^2 +\color{#3D99F6}{x-2}) \\ & = acx^6 + ax^5 + (bc-2a)x^4 + bx^3 + (c-2b)x^2 +x-2 \end{aligned}$

Equating coefficients, we have: $a=100$ , $b=10$ and $c=20$ . Therefore, we have:

$\begin{aligned} 2000x^6 + 100x^5 + 10x^3 + x - 2 & = 0 \\ (100x^4+10x^2 +1)(20x^2+x-2) & = 0 \end{aligned}$

$\implies \begin{cases} 100x^4+10x^2 +1 = 0 & \color{#D61F06}{\text{Discriminant }< 0 \text{, no real roots.}} \\ 20x^2+x-2 = 0 & \implies x = \dfrac {-1 \pm \sqrt{161}}{40} \end{cases}$

$\implies m + n + r = -1+161+40 = \boxed{200}$

Relevant wiki: Factoring Polynomials by Grouping

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

$\Rightarrow$ $2000x^{6} + \frac{x[(10x^{2})^{3} - 1]}{10x^{2} - 1} - 2 = 0$

$\Rightarrow$ $\frac{x(1000x^{6} - 1)}{10x^{2} - 1} = -2(1000x^{6} - 1)$

$\Rightarrow$ $(1000x^6 - 1) \left( \dfrac x{10x^2-1} + 2\right) = 0$

Hence, $1000x^{6} - 1 = 0$ or $\frac{x}{10x^{2} - 1} = -2$

$\Rightarrow$ $x^6 = \dfrac1{10^3}$ or $20x^{2} + x - 2 = 0$

$\Rightarrow$ $x^{2} = \frac{1}{10}$ (not required because we're looking for a different form of the root) or by quadratic formula $x = \frac{-1\pm \sqrt{161}}{40}$ ,

By solving the first equation, we cannot obtain a root in the form that was written in the question. So the only root we're looking for is $\dfrac{-1\pm\sqrt{161}}{40}$ .

Hence, $m = -1, n = 161, r = 40\Rightarrow m + n + r = \boxed{200}$ .

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

$2000x^6+100x^5+(200x^4-200x^4)+10x^3+(20x^2-20x^2)+x-2=0$

$(20x^2+x-2)(100x^4+10x^2+1)=0$

$20x^2+x-2=0$

$\therefore x=\dfrac{\sqrt{161}+(-1)}{40}$

$\therefore n+m+r=161-1+40=\boxed{200}$

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Note: There is a possibility that $100x^4+10x^2+1=0$ , but its roots are imaginary.

$2(1000x^6 - 1) + (x + x(10x^2) + x(10x^2)^2) = 0$

Let $L=x + x(10x^2) + x(10x^2)^2$

$\implies (10x^2)L=x(10x^2) + x(10x^2)^2 + x(10x^2)^3$ (since $x=0$ doesn't satisfy the solution, $10x^2 \neq 0$ )

$\implies (10x^2)L - L = x(10x^2)^3 - x$

$\implies L = x(\frac{1000x^6 - 1}{10x^2 - 1})$ , therefore:

$2(1000x^6 - 1) + x(\frac{1000x^6 - 1}{10x^2 - 1}) = 0$

$(1000x^6 - 1)(2 + \frac{x}{10x^2 - 1})= 0$

The second factor simpilifies to $20x^2 + x - 2=0$ , and is the only factor which returns a root of the form $\dfrac{m + \sqrt{n}}{r}$ , i. e. $\frac{-1+ \sqrt{161}}{40}$ .

Clearly, the required answer $m+n+r=(-1)+(161)+(40)=\boxed{200}$