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?
 A: Here's a technique that's sometimes called the '$ac$' method:
Quadratic functions have the form $$ax^{2} + bx + c.$$  In your case you have $$12x^{2} - 8x + 1,$$
so $a=12, b=-8$, and $c=1$.  The '$ac$' method says to do the following:
Step $1$: Look at $ac$ and $b$.  In your case $ac = 12\cdot 1 = 12$ and $b = -8$.
Step $2$:  Find two numbers that multiply to give you $ac$ and add to give you $b$.  In your case that would be $-2$ and $-6$ because $(-2)(-6) = 12 = ac$ and $-2 + (-6) = -8 = b$.
Step $3$: Rewrite your quadratic by splitting up the $bx$ part into your two numbers:
$$12x^{2} - 8x + 1 = 12x^{2} - 6x - 2x + 1.$$
Step $4$: Factor.
\begin{align*}
12x^{2} - 8x + 1 &= 12x^{2} - 6x - 2x + 1\\
&=6x(2x -1)  - (2x -1)\\
&=(2x-1)(6x-1).
\end{align*} 
A: Step 1. Which numbers multiply to give $12$? (Only positive cases needed to be considered here because the negative case ends up being equivalent to multiplying everything by $-1\times-1$)
$1, 12$ and 
$2, 6$ and 
$3, 4$
Step 2. Which numbers multiply to give $1$?
$1, 1$ and $-1, -1$
Step 3. Which numbers add to give $-8$?
From the top two lines or working you know the form will be $(ax+b)(cx+d)$ where $a, c$ are one of the pairs that multiplies to give $12$, and $b, d$ are one of the pairs that multiplies to give $1$.
From these numbers, you need $ad+bc = -8$. This means $b, d$ must be $-1$ (otherwise there's no way you can get -8, because the other numbers are positive.)
Hence you require $-a-c = -8$, so $a=2, c=6$.
Thus $12x^2−8x+1 = (2x-1)(6x-1)$
A: 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 $f(x) = 12(x-\frac{1}{2})(x-\frac{1}{6})=2(x-\frac{1}{2})\cdot6(x-\frac{1}{6})=(2x-1)(6x-1)$
A: There isn't a simple trick that always works; if you want something that works every time, the quadratic formula is the way to go. That being said, the rational root theorem says that the only possible rational roots to your polynomial are $\pm\frac1{r}$, where $r$ is a divisor of $12$. Also, the fact that the first degree term is the only one with negative sign means that both roots are positive.
Then you can get to checking. Finding one of the two roots $\frac12$ and $\frac16$ doesn't take too much time (there are only six candidates to check, and if you do then in order, then you hit on your second attempt). Once you have one root, Vieta's formulas give you the other.
If this brute forcing fails, then the roots are not rational, and the quadratic formula (or something equivalent) is the only way out. Also, if the constant term and second degree coefficients get too big, the rational root theorem gives too many candidates, so it stops being a viable option.
A: $$12x^2-8x+1=$$
$$12x^2-6x-2x+1=$$
$$6x (2x-1)-(2x-1)=$$
$$(2x-1)(6x-1) $$
or
$$12x^2-8x+1=$$
$$12x^2-3-8x+4=$$
$$3 (4x^2-1)-4(2x-1)= $$
$$3 (2x+1)(2x-1)-4 (2x-1)=$$
$$(2x-1)\left(3 (2x+1)-4\right)=$$
$$(2x-1)(6x+3-4) $$
A: When trying to factor something of the form $ax^2+bc+c$ where $a$ is neither $0$ or $1$, one common method is the $ac$-method.
The first step is to calculate the product $ac$ and to find a factor pair of $ac$ that adds up to $b$. In the case of $12x^2-8x+1$, notice that $ac =(12)(1) = 12$. Looking at factor pairs of $12$, it is not too difficult to see that $(-6)(-2) =12 = ac$ and that $(-6)+(-2) = -8 =b$.
The next step is to break the $bx$ term based on the factor pair we found. After that, we do what some call "factor by grouping" and this will allow us to get the final answer:
$$12x^2-8x+1 $$
$$= 12x^2-2x-6x+1$$
$$= (12x^2-2x) + (-6x+1)$$
$$= 2x(6x-1) - 1(6x-1)$$
$$= (6x-1)(2x-1).$$
A: Lets try making the first term a perfect square....then the rest may be easier.
$$\begin{align}
12x^2-8x+1 &= \frac{3\cdot12x^2-3\cdot8x+3\cdot1}{3}\\
&= \frac{36x^2-4\cdot 6x+3}{3}\\
&= \color{green}{\frac{(6x)^2-4(6x)+3}{3}}\\
&= \frac{(6x-1)(6x-3)}{3}\\
&= (6x-1)\frac{(6x-3)}{3}\\
&= (6x-1)(2x-1)\\
\end{align}$$
If that seems to confusing, try replacing $6x$ in the third step with an intermediary variable ($t$, for example), giving $\displaystyle\color{green}{\frac{t^2-4t+3}{3}}$.
A: Write $x=y/12$, so the polynomial becomes
$$
12\frac{y^2}{12^2}-\frac{8y}{12}+1=
\frac{1}{12}(y^2-8y+12)=\frac{1}{12}(y-2)(y-6)
$$
The decomposition of $y^2-8y+12$ is obtained from the search for two numbers whose sum is $8$ and product is $12$.
Substituting back $y=12x$, we have $y-2=2(6x-1)$ and $y-6=6(6x-1)$, so the denominator $12$ cancels out.
Note that in this way the coefficient of $y$ is always $1$. In general, for $a\ne0$, we have, setting $y=x/a$,
$$
ax^2+bx+c=a\frac{y^2}{a^2}+\frac{by}{a}+c=
\frac{1}{a}(y^2+by+ac)
$$
so the trick “sum and product” can be attempted at.
