$$2x^{2}-x+\frac{1}{8}-0$$
$\frac{\left(4x-1\right)^{2}}{8}$
$$2x^{2}-x+\frac{1}{8}+0$$
$$2x^{2}-x+\frac{1}{8}$$
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$$\frac{16x^{2}-8x+1}{8}$$
$$\left(4x-1\right)^{2}$$
$$\frac{\left(4x-1\right)^{2}}{8}$$
