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