How do you factor 2x^2+8x+42x2+8x+4?

2 Answers
Konstantinos Michailidis
Sep 16, 2015

It is 2x^2+8x+4=2*(x^2+4x+4)-4=2*(x+2)^2-2^2=(sqrt2(x+2))^2-2^2=(sqrt2(x+2)-2)*(sqrt2(x+2)+2)2x2+8x+4=2⋅(x2+4x+4)−4=2⋅(x+2)2−22=(√2(x+2))2−22=(√2(x+2)−2)⋅(√2(x+2)+2)

We used the formula a^2-b^2=(a-b)*(a+b)a2−b2=(a−b)⋅(a+b)

Alan P.
Sep 16, 2015

2x^2+8x+42x2+8x+4
color(white)("XXX")=color(green)(2)color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))XXX=2(x+2+√2)(x+2−√2)

Explanation:

Extract the obvious constant factor of 22
color(white)("XXX")color(green)(2)color(orange)((x^2+4x+2))XXX2(x2+4x+2)

Factor the second term using the quadratic formula for roots
color(white)("XXX")r=(-b+-sqrt(b^2-4ac))/(2a)XXXr=−b±√b2−4ac2a

In this case:
color(white)("XXX")r=(-4+-sqrt(4^2-4(1)(2)))/(2(1))XXXr=−4±√42−4(1)(2)2(1)

color(white)("XXX")= (-4+-sqrt(8))/2XXX=−4±√82

color(white)("XXX")=-2+-sqrt(2)XXX=−2±√2

If rr is a root then (x-r)(x−r) is a factor
So
color(white)("XXX")color(orange)((x^2+4x+2))XXX(x2+4x+2)
factors as
color(white)("XXX")=color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))XXX=(x+2+√2)(x+2−√2)

Giving the final factoring:
color(white)("XXX")2x^2+8x+4XXX2x2+8x+4
color(white)("XXX")=color(green)(2)color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))XXX=2(x+2+√2)(x+2−√2)

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