Factorize:

We have:

$2 \sqrt{3} x^{2}+x-5 \sqrt{3}$

We have to split 1 into two numbers such that their sum is 1 and product is 30 , i.e., $2 \sqrt{3} \times(-5 \sqrt{3})$.

Clearly, $6+(-5)=1$ and $6 \times(-5)=-30$

$\therefore 2 \sqrt{3} x^{2}+x-5 \sqrt{3}=2 \sqrt{3} x^{2}+6 x-5 x-5 \sqrt{3}$

$=2 \sqrt{3} x(x+\sqrt{3})-5(x+\sqrt{3})$

$=(x+\sqrt{3})(2 \sqrt{3} x-5)$