# How do you factor the expression 27x^2-3?

Apr 13, 2018

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

#### Explanation:

$\text{take out a "color(blue)"common factor } 3$

$= 3 \left(9 {x}^{2} - 1\right)$

$9 {x}^{2} - 1 \text{ is a "color(blue)"difference of squares}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$\text{here "9x^2=(3x)^2rArra=3x" and } b = 1$

$\Rightarrow 9 {x}^{2} - 1 = \left(3 x - 1\right) \left(3 x + 1\right)$

$\Rightarrow 27 {x}^{2} - 1 = 3 \left(3 x - 1\right) \left(3 x + 1\right)$

Apr 13, 2018

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

#### Explanation:

$27 {x}^{2} - 3$

$= 3 \left(9 {x}^{2} - 1\right)$

Differences of two squares rule:

• ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
• $= \left(\sqrt{{a}^{2}} + \sqrt{{b}^{2}}\right) \left(\sqrt{{a}^{2}} - \sqrt{{b}^{2}}\right)$

Because:
$\left(a + b\right) \left(a - b\right) = {a}^{2} + b a - a b - {b}^{2} = {a}^{2} - {b}^{2}$

${a}^{2}$ can be replaced with $9 {x}^{2}$

${b}^{2}$ can be replaced with $1$

$= 3 \left(\sqrt{9 {x}^{2}} + \sqrt{1}\right) \left(\sqrt{9 {x}^{2}} - \sqrt{1}\right)$

=color(red)(3(3x-1)(3x+1)