Factorise:

$2 x^{2}-x+\frac{1}{8}=2 x^{2}-\frac{1}{2} x-\frac{1}{2} x+\frac{1}{8}$

$=2 x\left(x-\frac{1}{4}\right)-\frac{1}{2}\left(x-\frac{1}{4}\right)$

$=\left(x-\frac{1}{4}\right)\left(2 x-\frac{1}{2}\right)$

Hence, factorisation of $2 x^{2}-x+\frac{1}{8}$ is $\left(x-\frac{1}{4}\right)\left(2 x-\frac{1}{2}\right)$.