# How do you factor x^2 - 4x - 2?

Feb 10, 2016

Use the quadratic formula to get:
$\textcolor{w h i t e}{\text{XXX}} \left(x - 2 + \sqrt{6}\right) \left(x - 2 - \sqrt{6}\right)$

#### Explanation:

$a {x}^{2} + b x + c$
the zeros of the expression are given by the formula
$\textcolor{w h i t e}{\text{XXX}} {x}_{z} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For the given expression
$\textcolor{w h i t e}{\text{XXX}} a = 1$
$\textcolor{w h i t e}{\text{XXX}} b = - 4$
$\textcolor{w h i t e}{\text{XXX}} c = - 2$

So
$\textcolor{w h i t e}{\text{XXX}} {x}_{z} = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} + 4 \left(1\right) \left(- 2\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{4 + \sqrt{16 + 8}}{2}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{4 \pm 2 \sqrt{6}}{2}$

$\textcolor{w h i t e}{\text{XXX}} = 2 \pm \sqrt{6}$

If ${x}_{z 1} \mathmr{and} {x}_{z 2}$ are zeros of a quadratic then the quadratic can be factored as:
$\textcolor{w h i t e}{\text{XXX}} \left(x - {x}_{z 1}\right) \left(x - {x}_{z 2}\right)$