Factorize:

We have :

$x^{2}-2 x+\frac{7}{16}$

$=\frac{16 x^{2}-32 x+7}{16}$

$=\frac{1}{16}\left(16 x^{2}-32 x+7\right)$

Now, we have to split $(-32)$ into two numbers such that their sum is $(-32)$ and their product is 112 , i.e., $16 \times 7$.

Clearly, $(-4)+(-28)=-32$ and $(-4) \times(-28)=112$.

$\therefore x^{2}-2 x+\frac{7}{16}=\frac{1}{16}\left(16 x^{2}-32 x+7\right)$

$=\frac{1}{16}\left(16 x^{2}-4 x-28 x+7\right)$

$=\frac{1}{16}[4 x(4 x-1)-7(4 x-1)]$

$=\frac{1}{16}(4 x-1)(4 x-7)$