Factorise $6x^{3} + 7x^{2} - 14x - 15$.

(N/A) Let $p(x) = 6x^{3} + 7x^{2} - 14x - 15$.
By the Factor Theorem,we test values for $x$. Let $x = -1$:
$p(-1) = 6(-1)^{3} + 7(-1)^{2} - 14(-1) - 15 = -6 + 7 + 14 - 15 = 0$.
Since $p(-1) = 0$,$(x + 1)$ is a factor.
Now,divide $6x^{3} + 7x^{2} - 14x - 15$ by $(x + 1)$ using long division or synthetic division:
$6x^{3} + 7x^{2} - 14x - 15 = (x + 1)(6x^{2} + x - 15)$.
Next,factorize the quadratic $6x^{2} + x - 15$ by splitting the middle term:
$6x^{2} + 10x - 9x - 15 = 2x(3x + 5) - 3(3x + 5) = (2x - 3)(3x + 5)$.
Thus,the complete factorization is $(x + 1)(3x + 5)(2x - 3)$.