Question:
Find the value
$x^{2}-\sqrt{3} x-6$
Solution:
Splitting the middle term,
$=x^{2}-2 \sqrt{3} x+\sqrt{3} x-6$
$[\therefore-\sqrt{3}=-2 \sqrt{3}+\sqrt{3}$ also $-2 \sqrt{3} \times \sqrt{3}=-6]$
$=x(x-2 \sqrt{3})+\sqrt{3}(x-2 \sqrt{3})$
$=(x-2 \sqrt{3})(x+\sqrt{3})$
$\therefore x^{2}-\sqrt{3} x-6=(x-2 \sqrt{3})(x+\sqrt{3})$