Factorize:
$(2 x-3)^{2}-8 x+12$
We have:
$(2 x-3)^{2}-8 x+12=(2 x-3)^{2}-4(2 x-3)$
$=(2 x-3)[(2 x-3)-4]$
$=(2 x-3)(2 x-3-4)$
$=(2 x-3)(2 x-7)$
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