How to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without quadratic equation? After almost seven years I recently started again learning math and have few holes in my algebra knowledge, so I apologize for the beginner question.
My question is:
Is there any simple trick to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without using quadratic equation?
And if it's possible, what is step by step procedure?

Not sure what you mean by "without quadratic equation" but I'll assume it means without using the quadratic formula. So we can factorise it by completing the square:
Let $f(x) = 12x^2-8x+1$, then:
$f(x) = 12(x^2-\frac{8}{12}x+\frac{1}{12})=12(x^2-\frac{2}{3}x+\frac{1}{12})$
$f(x) = 12[(x-\frac{1}{2}\cdot\frac{2}{3})^2-(\frac{1}{3})^2+\frac{1}{12}]$
$f(x) = 12[(x-\frac{1}{3})^2-\frac{1}{9}+\frac{1}{12}]$
$f(x) = 12[(x-\frac{1}{3})^2+\frac{3-4}{36}]=12[(x-\frac{1}{3})^2-\frac{1}{36}]$
So then we set $f(x) = 0:$
$(x-\frac{1}{3})^2-\frac{1}{36}=0\implies x=\frac{1}{3}\pm\frac{1}{6}\implies x=\frac{1}{2}\quad\text{or}\quad x=\frac{1}{6}$
Therefore $f(x) = 12(x-\frac{1}{2})(x-\frac{1}{6})=2(x-\frac{1}{2})\cdot6(x-\frac{1}{6})=(2x-1)(6x-1)$
