Factorise :

$x^2 - 2x - 9$.

Given:

$x^2 - 2x - 9$

Using the quadratic formula:

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the value of a = 1, b = -2 and c = -9,

$\text{x} = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 1 \times (-9)}}{2 \times 1}\\[1em] \text{x} = \dfrac{2 \pm \sqrt{4 + 36}}{2}\\[1em] \text{x} = \dfrac{2 \pm \sqrt{40}}{2}\\[1em] \text{x} = \dfrac{2 \pm 2\sqrt{10}}{2}\\[1em] \text{x} = 1 \pm \sqrt{10}\\[1em] \text{x} - 1 + \sqrt{10} = 0 \text{ or } \text{x} - 1 - \sqrt{10} = 0$

Hence, $x^2 - 2x - 9 = (x - 1 + \sqrt{10})(x - 1 - \sqrt{10})$.