Question

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??

Answer 1

We know that if `P(alpha)=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+4x+2=0`

`x=(-4+-sqrt(16-8))/2=(-4+-sqrt8)/2`

But `sqrt8=sqrt(4·2)=2sqrt2`, then `x=(-4+-2sqrt2)/2` and extracting common factor 2, we have

`x=-2+-sqrt2`. That means `-2+sqrt2` and `-2-sqrt2` are roots of initial polynomial, and for statement of beginning

`x^2+4x+2=(x+2+sqrt2)(x+2-sqrt2)`

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

Hope this helps

Answer 2

`"see explanation"`

Explanation:

`"given "x=a" is a root of a polynomial then"`

`(x-a)" is a factor of the polynomial"`

`"find the roots using the "color(blue)"quadratic formula"`

`"with "a=1,b=4" and "c=2`

`x=(-4+-sqrt(16-8))/2=-2+-sqrt2`

`"thus factors are"`

`(x-(-2-sqrt2))" and "(x-(-2+sqrt2))`

`(x+sqrt2+2)" and "(x-sqrt2+2)`

`rArrx^2+4x+2=(x+sqrt2+2)(x-sqrt2+2)`

Answer 3

Kindly refer to Explanation.

Explanation:

Completing the square of the quadr. poly. `x^2+4x+2`,

`(x^2+4x)+2=(x^2+4x+4)-4+2=`,

`=(x+2)^2-2`,

`=(x+2)^2-(sqrt2)^2`,

`=(x+2+sqrt2)(x+2-sqrt2)`, as desired!