How do you factor 8x^3 - 12x^2 - 2x + 3 ?

Feb 1, 2017

$8 {x}^{3} - 12 {x}^{2} - 2 x + 3 = \left(2 x - 1\right) \left(2 x + 1\right) \left(2 x - 3\right)$

Explanation:

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping.

We will also use the difference of squares identity:

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

with $a = \left(2 x - 1\right)$ and $b = 1$ as follows:

$8 {x}^{3} - 12 {x}^{2} - 2 x + 3 = \left(8 {x}^{3} - 12 {x}^{2}\right) - \left(2 x - 3\right)$

$\textcolor{w h i t e}{8 {x}^{3} - 12 {x}^{2} - 2 x + 3} = 4 {x}^{2} \left(2 x - 3\right) - 1 \left(2 x - 3\right)$

$\textcolor{w h i t e}{8 {x}^{3} - 12 {x}^{2} - 2 x + 3} = \left(4 {x}^{2} - 1\right) \left(2 x - 3\right)$

$\textcolor{w h i t e}{8 {x}^{3} - 12 {x}^{2} - 2 x + 3} = \left({\left(2 x\right)}^{2} - {1}^{2}\right) \left(2 x - 3\right)$

$\textcolor{w h i t e}{8 {x}^{3} - 12 {x}^{2} - 2 x + 3} = \left(2 x - 1\right) \left(2 x + 1\right) \left(2 x - 3\right)$