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

Aug 18, 2016

$9 {x}^{3} - 9 {x}^{2} - 4 x + 4 = \left(3 x - 2\right) \left(3 x + 2\right) \left(x - 1\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

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

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

$= \left(9 {x}^{3} - 9 {x}^{2}\right) - \left(4 x - 4\right)$

$= 9 {x}^{2} \left(x - 1\right) - 4 \left(x - 1\right)$

$= \left(9 {x}^{2} - 4\right) \left(x - 1\right)$

$= \left({\left(3 x\right)}^{2} - {2}^{2}\right) \left(x - 1\right)$

$= \left(3 x - 2\right) \left(3 x + 2\right) \left(x - 1\right)$