# How do you factor 2x^2-5x+1?

Mar 2, 2017

$2 {x}^{2} - 5 x + 1 = 2 \left(x - \frac{5 + \sqrt{17}}{4}\right) \left(x - \frac{5 - \sqrt{17}}{4}\right)$

#### Explanation:

$2 {x}^{2} - 5 x + 1 = 2 \left({x}^{2} - \frac{5}{2} x + \frac{1}{2}\right)$

Consider ${x}^{2} - \frac{5}{2} x + \frac{1}{2} = 0$ and solve by completing a square
${x}^{2} - \frac{5}{2} x + \frac{1}{2} = 0$

${\left(x - \frac{5}{4}\right)}^{2} - {\left(\frac{5}{4}\right)}^{2} + \frac{1}{2} = 0$

${\left(x - \frac{5}{4}\right)}^{2} - \frac{25}{16} + \frac{1}{2} = 0$

${\left(x - \frac{5}{4}\right)}^{2} - \frac{17}{16} = 0$

${\left(x - \frac{5}{4}\right)}^{2} = \frac{17}{16}$

$\left(x - \frac{5}{4}\right) = \pm \sqrt{\frac{17}{16}} = \pm \frac{\sqrt{17}}{4}$

$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4} = \frac{5 \pm \sqrt{17}}{4}$

$\left({x}^{2} - \frac{5}{2} x + \frac{1}{2}\right) = \left(x - \frac{5 + \sqrt{17}}{4}\right) \left(x - \frac{5 - \sqrt{17}}{4}\right)$

therefore

$2 {x}^{2} - 5 x + 1 = 2 \left(x - \frac{5 + \sqrt{17}}{4}\right) \left(x - \frac{5 - \sqrt{17}}{4}\right)$