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??

3 Answers
Jun 9, 2018

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

Jun 9, 2018

"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)

Jun 9, 2018

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!