# How do you factor completely 2a^2 - 4a - 2?

Oct 28, 2016

$2 {a}^{2} - 4 a - 2 = 2 \left(a - 1 - \sqrt{3}\right) \left(a - 1 + \sqrt{3}\right)$

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

Use this with $A = \left(a - 1\right)$ and $B = \sqrt{3}$ as follows:

$2 {a}^{2} - 4 a - 2 = 2 \left({a}^{2} - 2 a - 2\right)$

$\textcolor{w h i t e}{2 {a}^{2} - 4 a - 2} = 2 \left({a}^{2} - 2 a + 1 - 3\right)$

$\textcolor{w h i t e}{2 {a}^{2} - 4 a - 2} = 2 \left({\left(a - 1\right)}^{2} - {\left(\sqrt{3}\right)}^{2}\right)$

$\textcolor{w h i t e}{2 {a}^{2} - 4 a - 2} = 2 \left(\left(a - 1\right) - \sqrt{3}\right) \left(\left(a - 1\right) + \sqrt{3}\right)$

$\textcolor{w h i t e}{2 {a}^{2} - 4 a - 2} = 2 \left(a - 1 - \sqrt{3}\right) \left(a - 1 + \sqrt{3}\right)$