First, solve $\displaystyle 2x-1$: $\displaystyle 2x-1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$

Then substitute $\displaystyle \left(\frac{1}{2}\right)$ for $\displaystyle x$ in $\displaystyle 2x^4-x^3-6x^2+5x-1$:

$\displaystyle 2\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^3-6\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\rig ht)-1$ = $\displaystyle \frac{1}{8}-\frac{1}{8}-\frac{3}{2}+\frac{5}{2} = 0$

Thus, by the factor theorem, $\displaystyle 2x-1$ is a factor of $\displaystyle 2x^4-x^3-6x^2+5x-1$.