# How do you factor completely 16y^3 - 14y^2 - 12y?

Feb 18, 2017

$16 {y}^{3} - 14 {y}^{2} - 12 y = 16 y \left(y - \frac{7}{16} - \frac{\sqrt{241}}{16}\right) \left(y - \frac{7}{16} + \frac{\sqrt{241}}{16}\right)$

#### Explanation:

Complete the square then use the difference of squares identity:

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

with $a = \left(16 y - 7\right)$ and $b = \sqrt{241}$ as follows:

$16 {y}^{3} - 14 {y}^{2} - 12 y = \frac{1}{16} y \left(256 {y}^{2} - 224 y - 192\right)$

$\textcolor{w h i t e}{16 {y}^{3} - 14 {y}^{2} - 12 y} = \frac{1}{16} y \left({\left(16 y\right)}^{2} - 2 \left(16 y\right) \left(7\right) + {\left(7\right)}^{2} - 241\right)$

$\textcolor{w h i t e}{16 {y}^{3} - 14 {y}^{2} - 12 y} = \frac{1}{16} y \left({\left(16 y - 7\right)}^{2} - {\left(\sqrt{241}\right)}^{2}\right)$

$\textcolor{w h i t e}{16 {y}^{3} - 14 {y}^{2} - 12 y} = \frac{1}{16} y \left(16 y - 7 - \sqrt{241}\right) \left(16 y - 7 + \sqrt{241}\right)$

$\textcolor{w h i t e}{16 {y}^{3} - 14 {y}^{2} - 12 y} = 16 y \left(y - \frac{7}{16} - \frac{\sqrt{241}}{16}\right) \left(y - \frac{7}{16} + \frac{\sqrt{241}}{16}\right)$