In order to factorise $2 x^{3}-7 x-15$, we have to find two numbers $p$ and $q$ such that

$p+q=-7$ and $p q=-30.$

Clearly, $(-10)+3=-7$ and $(-10) \times 3=-30 .$

So, we write the middle term $-7 x$ as $(-10 x)+3 x$.

$\therefore 2 x^{2}-7 x-15=2 x^{2}-10 x+3 x-15.$

$\therefore 2 x^{2}-7 x-15=2 x^{2}-10 x+3 x-15$

$=2 x(x-5)+3(x-5)$

$(x-5)(2 x+3)$