<s>[INST] Is there any formula for $16x^2 - 8x + 1$ to $(4x-5)(4x+3)$? I need to turn this $16x^2 - 8x + 1$ to $(4x-5)(4x+3)$ and I completely lost my logic on this one.
When trying to get $D$ with $D=b^2-4ac$ formula, I get $D < 0$.
Can someone explain me how to get these $-5$ and $+3$. Should I do this in my mind or is there another formula for this?
Thank you.
 [/INST] Hint: Completing the square you have 
$$16\Big[x^2 - \frac{1}{2}x\Big]  +1  = 16\Big[\left(x - \frac{1}{4}\right)^2 - \frac{1}{16}\Big]  +1 = 16\left(x - \frac{1}{4}\right)^2 - 1  +1 = (4x - 1)^2$$
Now think of $(4x -1)^2 - 4^2$ and use $a^2-b^2 = (a - b)(a+b)$. 
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