# How do you factor 24x^3-6x^2+8x-2?

Nov 15, 2016

$24 {x}^{3} - 6 {x}^{2} + 8 x - 2 = 2 \left(3 {x}^{2} + 1\right) \left(4 x - 1\right)$

#### Explanation:

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

$24 {x}^{3} - 6 {x}^{2} + 8 x - 2 = \left(24 {x}^{3} - 6 {x}^{2}\right) + \left(8 x - 2\right)$

$\textcolor{w h i t e}{24 {x}^{3} - 6 {x}^{2} + 8 x - 2} = 6 {x}^{2} \left(4 x - 1\right) + 2 \left(4 x - 1\right)$

$\textcolor{w h i t e}{24 {x}^{3} - 6 {x}^{2} + 8 x - 2} = \left(6 {x}^{2} + 2\right) \left(4 x - 1\right)$

$\textcolor{w h i t e}{24 {x}^{3} - 6 {x}^{2} + 8 x - 2} = 2 \left(3 {x}^{2} + 1\right) \left(4 x - 1\right)$

The remaining quadratic factor has no linear factors with Real coefficients, since $3 {x}^{2} + 1 \ge 1$ for all Real values of $x$.