Rational Root Theorem won't save you this time

Due to me wanting to get the checkpoint of 1000 upvotes on solutions, I will post one now.

We want to use Rational Root Theorem, but that $\sqrt{6}$ prevents us from doing so! How can we get rid of it?

If we can somehow divide the entire equation by $\sqrt{6}$ without the $x^3$ and $x$ coefficients obtaining a $\sqrt{6}$ , then we could use RRT. But how can we do that?

Here is the trick: notice that $x^3$ and $x$ are both odd-degree terms. if we plugged in $P(x'\sqrt{6})$ , then our polynomial becomes $3(x'\sqrt{6})^3-17(x'\sqrt{6})+5\sqrt{6}=\sqrt{6}(18x'^3-17x'+5)=0$

Now we can divide by $\sqrt{6}$ on both sides! We obtain the polynomial $P(x'\sqrt{6})=18x'^3-17x'+5$

Now we use Rational Root Theorem and a bit of luck to find that $x'=\dfrac{1}{3}$ is a root of this new polynomial. Using polynomial division to divide $18x'^3-17x'+5$ by $3x'-1$ , we obtain the quadratic $6x'^2+2x'-5$

The roots of this quadratic can be found by the Quadratic Formula; skipping over the details, we obtain that $x'=\dfrac{-1\pm \sqrt{31}}{6}$

Now that we have all three roots of the modified quadratic, we need to find the roots of our original quadratic. Remember that we substituted $x=x'\sqrt{6}$ . Thus, to get our original roots, we must multiply the roots we have by $\sqrt{6}$ . Thus, our roots are: $x=\dfrac{\sqrt{6}}{3}$ $x=\dfrac{-\sqrt{6}\pm \sqrt{186}}{6}$

The smallest root in these three is obviously $\dfrac{-\sqrt{6}-\sqrt{186}}{6}$ since it's the only negative root. Thus, we just need to find the value of $r_1+r_2=\dfrac{\sqrt{6}}{3}+\dfrac{-\sqrt{6}+\sqrt{186}}{6}=\dfrac{\sqrt{6}+\sqrt{186}}{6}$

so our final answer is $186+6+6=\boxed{198}$ .