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

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Answer 1 · 2016-08-18T18:32:26

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

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:

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

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

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

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

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

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

How do you factor completely 9x^3-9x^2-4x+49x^3-9x^2-4x+4?