Factorise $: x^{3}-x^{2}-17 x-15$

The sum of the coefficients of odd power

terms of $x=1-17=-16$.

The sum of the coefficients of the even power

terms of $x=-1-15=-16$.

Hence, $(x+1)$ is a factor of $p(x)$

$x^{3}-x^{2}-17 x-15$

$=\underline{x^{3}+x^{2}}-\underline{2 x^{2}-2 x}-\underline{15 x-15}$

[Splitting the terms to get $x+1$ as a factor]

$=x^{2}(x+1)-2 x(x+1)-15(x+1)$

$=(x+1)\left(x^{2}-2 x-15\right)$

$=(x+1)\left(x^{2}-5 x+3 x-15\right)$

[By splitting the middle term]

$=(x+1)[x(x-5)+3(x-5)]$

$=(x+1)(x-5)(x+3)$