# Is it possible to factor y= x^2 -x - 11 ? If so, what are the factors?

Dec 22, 2015

Yes, with irrational coefficients:

${x}^{2} - x - 11 = \left(x - \frac{1}{2} - \frac{3 \sqrt{5}}{2}\right) \left(x - \frac{1}{2} + \frac{3 \sqrt{5}}{2}\right)$

#### Explanation:

Complete the square and use the difference of squares identity:

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

with $a = \left(x - \frac{1}{2}\right)$ and $b = \frac{\sqrt{45}}{2}$ ...

${x}^{2} - x - 11$

$= {x}^{2} - x + \frac{1}{4} - \frac{45}{4}$

$= {\left(x - \frac{1}{2}\right)}^{2} - \frac{45}{4}$

$= {\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{45}}{2}\right)}^{2}$

$= \left(\left(x - \frac{1}{2}\right) - \frac{\sqrt{45}}{2}\right) \left(\left(x - \frac{1}{2}\right) + \frac{\sqrt{45}}{2}\right)$

$= \left(x - \frac{1}{2} - \frac{\sqrt{45}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{45}}{2}\right)$

Finally note that if $a , b \ge 0$ then $\sqrt{a b} = \sqrt{a} \sqrt{b}$, so we can simplify:

$\sqrt{45} = \sqrt{9 \cdot 5} = \sqrt{9} \cdot \sqrt{5} = 3 \sqrt{5}$

So we can write:

${x}^{2} - x - 11 = \left(x - \frac{1}{2} - \frac{3 \sqrt{5}}{2}\right) \left(x - \frac{1}{2} + \frac{3 \sqrt{5}}{2}\right)$