# How do you factor 9x^2 + 6x + 1 = 12x?

May 4, 2018

$x = \frac{1}{3}$

#### Explanation:

$9 {x}^{2} + 6 x + 1 = 12 x$

subtract $12 x$ from both sides

$9 {x}^{2} + 6 x + 1 - 12 x = 0$

$\textcolor{red}{\text{6x - 12x}}$

$9 {x}^{2} \textcolor{red}{\text{- 6x}} + 1 = 0$

then my way that I use when the coefficient of ${x}^{2}$ not $1$
that I multiply the constant by the coefficient of ${x}^{2}$ then I try to find two numbers that their product equals to the coefficient of ${x}^{2}$ and their sum equals to the coefficient of $x$.

like here for example

$9 {x}^{2} - 6 x + 1 = 0$

$9 \cdot 1 = 9$

so the two numbers are $- 3$ and $- 3$

their product = 9 and their sum = -6

so then I put them in $\textcolor{red}{\text{-6x}}$ place
$9 {x}^{2} \textcolor{red}{\text{- 3x - 3x}} + 1 = 0$

the by grouping take the common factor

$\textcolor{red}{\text{9x^2 - 3x" color(blue)" - 3x + 1= 0}}$

$3 x \left(3 x - 1\right) - \left(3 x - 1\right) = 0$

so

$\left(3 x - 1\right) \left(3 x - 1\right) = 0$

${\left(3 x - 1\right)}^{2} = 0$

square root both sides

$3 x - 1 = 0$

add 1 to both sides

$3 x = 1$

divide both sides by 3

$x = \frac{1}{3}$