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

2 Answers

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)#

We used the formula #a^2-b^2=(a-b)*(a+b)#

Sep 16, 2015

Answer:

#2x^2+8x+4#
#color(white)("XXX")=color(green)(2)color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))#

Explanation:

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

Factor the second term using the quadratic formula for roots
#color(white)("XXX")r=(-b+-sqrt(b^2-4ac))/(2a)#

In this case:
#color(white)("XXX")r=(-4+-sqrt(4^2-4(1)(2)))/(2(1))#

#color(white)("XXX")= (-4+-sqrt(8))/2#

#color(white)("XXX")=-2+-sqrt(2)#

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

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