# How do you factor 2x^2+8x+4?

It is $2 {x}^{2} + 8 x + 4 = 2 \cdot \left({x}^{2} + 4 x + 4\right) - 4 = 2 \cdot {\left(x + 2\right)}^{2} - {2}^{2} = {\left(\sqrt{2} \left(x + 2\right)\right)}^{2} - {2}^{2} = \left(\sqrt{2} \left(x + 2\right) - 2\right) \cdot \left(\sqrt{2} \left(x + 2\right) + 2\right)$

We used the formula ${a}^{2} - {b}^{2} = \left(a - b\right) \cdot \left(a + b\right)$

Sep 16, 2015

$2 {x}^{2} + 8 x + 4$
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{g r e e n}{2} \textcolor{red}{\left(x + 2 + \sqrt{2}\right)} \textcolor{b l u e}{\left(x + 2 - \sqrt{2}\right)}$

#### Explanation:

Extract the obvious constant factor of $2$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{2} \textcolor{\mathmr{and} a n \ge}{\left({x}^{2} + 4 x + 2\right)}$

Factor the second term using the quadratic formula for roots
$\textcolor{w h i t e}{\text{XXX}} r = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case:
$\textcolor{w h i t e}{\text{XXX}} r = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(1\right) \left(2\right)}}{2 \left(1\right)}$

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

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

If $r$ is a root then $\left(x - r\right)$ is a factor
So
$\textcolor{w h i t e}{\text{XXX}} \textcolor{\mathmr{and} a n \ge}{\left({x}^{2} + 4 x + 2\right)}$
factors as
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{\left(x + 2 + \sqrt{2}\right)} \textcolor{b l u e}{\left(x + 2 - \sqrt{2}\right)}$

Giving the final factoring:
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} + 8 x + 4$
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{g r e e n}{2} \textcolor{red}{\left(x + 2 + \sqrt{2}\right)} \textcolor{b l u e}{\left(x + 2 - \sqrt{2}\right)}$