Value Plugging?

Completing square, we want to evaluate the expression at $x=4-\sqrt 3$ . We first show that $4-\sqrt 3$ is a root of the quartic $f(x)=x^4-6x^3-2x^2+18x+13.$ Indeed, applying Rational Root Theorem suggestions twice enables us to find a root $-1$ of multiplicity $2$ (since $13$ is a prime, we have few cases to check). Now factorizing gives $f(x)=x^4-6x^3-2x^2+18x+13=\left(x+1\right)^2 \left(x^2-8x+13\right).$ Bashing the quadratic, call it $g(x)$ , we see that $x=4-\sqrt 3$ is indeed a root of $g$ and $f$ . Now since $\dfrac{x^4-6x^3-2x^2+18x+1286}{x^2-8x+32} = \dfrac{f(x)+1273}{g(x)+19}$ and at $x=4-\sqrt 3$ we have $f(x)=g(x)=0$ , we finally get $\dfrac{f(x)+1273}{g(x)+19}=\dfrac{1273}{19}=\boxed{67}$