Is it possible to factor #y=2x^2 + 8x+4#? If so, what are the factors?

2 Answers
MeneerNask
Dec 8, 2015

First you can take out the factor #2#

Explanation:

#->y=2(x^2+4x+2)#

This cannot be factored any further without using radical expressions.

George C.
Dec 8, 2015

#2x^2+8x+4 = 2(x^2+4x+2)#

#= 2(x+2+sqrt(2))(x+2-sqrt(2))#

Explanation:

Given #y = 2x^2+8x+4# first separate out the common scalar factor #2# to get:

#y = 2(x^2+4x+2)#

The quadratic factor is in the form #ax^2+bx+c# with #a=1#, #b=4#, #c=2#. This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-4+-sqrt((-4)^2-(4xx1xx2)))/(2xx1)#

#=(-4+-sqrt(8))/2 = (-4+-sqrt(2^2*2))/2 = (-4+-2sqrt(2))/2 = -2+-sqrt(2)#

These zeros correspond to factors #(x+2+sqrt(2))# and #(x+2-sqrt(2))#

#color(white)()#
Another way of finding this is by completing the square, then using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = (x+2)# and #b = sqrt(2)# as follows:

#y = 2x^2+8x+4#

#=2(x^2+4x+2)#

#=2(x^2+4x+4-2)#

#=2((x+2)^2-2)#

#=2((x+2)^2-(sqrt(2))^2)#

#=2((x+2)-sqrt(2))((x+2)+sqrt(2))#

#=2(x+2-sqrt(2))(x+2+sqrt(2))#

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