# How do you factor 8x^3-4x^2-2x+1?

Mar 26, 2016

$8 {x}^{3} - 4 {x}^{2} - 2 x + 1 = \left(2 x - 1\right) \left(2 x + 1\right) \left(2 x - 1\right)$

#### Explanation:

Factor by grouping and using the difference of squares identity:

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

with $a = 2 x$ and $b = 1$ as follows:

$8 {x}^{3} - 4 {x}^{2} - 2 x + 1$

$= \left(8 {x}^{3} - 4 {x}^{2}\right) - \left(2 x - 1\right)$

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

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

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

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