# Can any anyone explain how x^2+4x+2 become (x-sqrt2+2)(x+sqrt2+2) x^2+4x+2=(x-sqrt2+2)(x+sqrt2+2) helppppp??

Jun 9, 2018

We know that if $P \left(\alpha\right) = 0$ then $x - \alpha$ is a factor of polynomial. That means: $\alpha$ is a root of polynomial

In order to factorize a degree two polynomial, we can use the quadratic formula suposing that ${x}^{2} + 4 x + 2 = 0$

$x = \frac{- 4 \pm \sqrt{16 - 8}}{2} = \frac{- 4 \pm \sqrt{8}}{2}$

But sqrt8=sqrt(4·2)=2sqrt2, then $x = \frac{- 4 \pm 2 \sqrt{2}}{2}$ and extracting common factor 2, we have

$x = - 2 \pm \sqrt{2}$. That means $- 2 + \sqrt{2}$ and $- 2 - \sqrt{2}$ are roots of initial polynomial, and for statement of beginning

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

To check the solution we can multiply both factors and the result must be the initial polynomial

Hope this helps

Jun 9, 2018

$\text{see explanation}$

#### Explanation:

$\text{given "x=a" is a root of a polynomial then}$

$\left(x - a\right) \text{ is a factor of the polynomial}$

$\text{find the roots using the "color(blue)"quadratic formula}$

$\text{with "a=1,b=4" and } c = 2$

$x = \frac{- 4 \pm \sqrt{16 - 8}}{2} = - 2 \pm \sqrt{2}$

$\text{thus factors are}$

$\left(x - \left(- 2 - \sqrt{2}\right)\right) \text{ and } \left(x - \left(- 2 + \sqrt{2}\right)\right)$

$\left(x + \sqrt{2} + 2\right) \text{ and } \left(x - \sqrt{2} + 2\right)$

$\Rightarrow {x}^{2} + 4 x + 2 = \left(x + \sqrt{2} + 2\right) \left(x - \sqrt{2} + 2\right)$

Jun 9, 2018

Kindly refer to Explanation.

#### Explanation:

Completing the square of the quadr. poly. ${x}^{2} + 4 x + 2$,

$\left({x}^{2} + 4 x\right) + 2 = \left({x}^{2} + 4 x + 4\right) - 4 + 2 =$,

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

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

$= \left(x + 2 + \sqrt{2}\right) \left(x + 2 - \sqrt{2}\right)$, as desired!