How do you factor completely #9x^3-9x^2-4x+4#?

1 Answer
George C.
Aug 18, 2016

#9x^3-9x^2-4x+4=(3x-2)(3x+2)(x-1)#

Explanation:

The difference of squares identity can be written:

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

We use this with #a=3x# and #b=2#, but first factor by grouping:

#9x^3-9x^2-4x+4#

#=(9x^3-9x^2)-(4x-4)#

#=9x^2(x-1)-4(x-1)#

#=(9x^2-4)(x-1)#

#=((3x)^2-2^2)(x-1)#

#=(3x-2)(3x+2)(x-1)#

Related questions
See all questions in Factoring Completely
Impact of this question
2065 views around the world
You can reuse this answer
Creative Commons License