How do you factor 6x^3-6x^2-x+1?

Mar 23, 2018

$6 {x}^{3} - 6 {x}^{2} - x + 1$
$= x \left(6 {x}^{2} - 6 x - 1 + 1\right)$
$= x \left(6 {x}^{2} - 6 x\right)$
$= 6 x \left({x}^{2} - x\right)$

Mar 23, 2018

$6 {x}^{3} - 6 {x}^{2} - x + 1 = \left(6 {x}^{2} - 1\right) \left(x - 1\right)$

$\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left(\sqrt{6} x - 1\right) \left(\sqrt{6} x + 1\right) \left(x - 1\right)$

Explanation:

This cubic quadrinomial factors by grouping and using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = \sqrt{6} x$ and $B = 1$ as follows:

$6 {x}^{3} - 6 {x}^{2} - x + 1 = \left(6 {x}^{3} - 6 {x}^{2}\right) - \left(x - 1\right)$

$\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = 6 {x}^{2} \left(x - 1\right) - 1 \left(x - 1\right)$

$\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left(6 {x}^{2} - 1\right) \left(x - 1\right)$

$\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left({\left(\sqrt{6} x\right)}^{2} - {1}^{2}\right) \left(x - 1\right)$

$\textcolor{w h i t e}{6 {x}^{3} - 6 {x}^{2} - x + 1} = \left(\sqrt{6} x - 1\right) \left(\sqrt{6} x + 1\right) \left(x - 1\right)$