<s>[INST] 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?
 [/INST] 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 $f(x) = 12(x-\frac{1}{2})(x-\frac{1}{6})=2(x-\frac{1}{2})\cdot6(x-\frac{1}{6})=(2x-1)(6x-1)$
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