How do you factor #3a^2 + 4a - 4#?

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Answer 1 · 2016-04-16T07:02:36

#3a^2+4a-4 = (3a-2)(a+2)#

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=(3a+2)# and #B=4#, as follows:

To make the arithmetic a little easier, first multiply by #3# (making the leading term into a square), remembering to divide by #3# at the end:

#3(3a^2+4a-4)#

#=9a^2+12a-12#

#=(3a+2)^2-4-12#

#=(3a+2)^2-4^2#

#=((3a+2)-4)((3a+2)+4)#

#=(3a-2)(3a+6)#

#=3(3a-2)(a+2)#

So dividing by #3# we have:

#3a^2+4a-4 = (3a-2)(a+2)#