# Question #47c74

Jan 25, 2018

$3 \left(x + 1\right) \left(x - 1\right)$

#### Explanation:

Factor out a 3 from each term in the equation:

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

"Pull" (reverse-distribute) the 3:

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

${x}^{2} - 1$ = $\left(x + 1\right) \left(x - 1\right)$ because it's a difference of squares:

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

Here's how differences of squares work:

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

This is true for any variable x and a $\left(x + a\right) \left(x - a\right)$
$\left(x + a\right) \left(x - a\right)$
$= \left(x \left(x - a\right)\right) + a \left(x - a\right)$
$= {x}^{2} - a x + a x - {a}^{2}$
$= {x}^{2} - {a}^{2}$