# How do you factor y= x^2-3x-1 ?

Dec 24, 2015

It requires irrational coefficients, but...

Complete the square and use the difference of squares identity to find:

$y = {x}^{2} - 3 x - 1$

$= \left(x - \frac{3}{2} - \frac{\sqrt{13}}{2}\right) \left(x - \frac{3}{2} + \frac{\sqrt{13}}{2}\right)$

#### Explanation:

Note that:

${\left(x - \frac{3}{2}\right)}^{2} = {x}^{2} - 3 x + \frac{9}{4}$

Also, the difference of squares identity can be written:

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

We use this with $a = \left(x - \frac{3}{2}\right)$ and $b = \frac{\sqrt{13}}{2}$ below...

$y = {x}^{2} - 3 x - 1$

$= {x}^{2} - 3 x + \frac{9}{4} - \frac{9}{4} - 1$

$= {\left(x - \frac{3}{2}\right)}^{2} - \frac{13}{4}$

$= {\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{13}}{2}\right)}^{2}$

$= \left(\left(x - \frac{3}{2}\right) - \frac{\sqrt{13}}{2}\right) \left(\left(x - \frac{3}{2}\right) + \frac{\sqrt{13}}{2}\right)$

$= \left(x - \frac{3}{2} - \frac{\sqrt{13}}{2}\right) \left(x - \frac{3}{2} + \frac{\sqrt{13}}{2}\right)$