How do you factor #12x^2+8x-16#?

1 Answer
Aug 22, 2017

#12x^2+8x-16 = 4/3(3x+1-sqrt(13))(3x+1+sqrt(13))#

Explanation:

Given:

#12x^2+8x-16#

We can factor this by completing the square then using the difference of squares identity:

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

with #a=(3x+1)# and #b=sqrt(13)# as follows.

First note that all of the terms are divisible by #4# and that in order to make the leading coefficient into a square number without making the other coefficients into fractions, we need to multiply by #3#. So for simplicity, start by multiplying by #3/4#...

#3/4(12x^2+8x-16) = 9x^2+6x-12#

#color(white)(3/4(12x^2+8x-16)) = (3x)^2+2(3x)+1-13#

#color(white)(3/4(12x^2+8x-16)) = (3x+1)^2-(sqrt(13))^2#

#color(white)(3/4(12x^2+8x-16)) = ((3x+1)-sqrt(13))((3x+1)+sqrt(13))#

#color(white)(3/4(12x^2+8x-16)) = (3x+1-sqrt(13))(3x+1+sqrt(13))#

Then multiplying both ends by #4/3#, we find:

#12x^2+8x-16 = 4/3(3x+1-sqrt(13))(3x+1+sqrt(13))#