Factorise $: x^{3}+x^{2}-26 x+24$

Sum of all the coefficients of terms of the polynomial

$=1+1-26+24$

$=26-26=0$

So, $(x-1)$ is a factor of $p(x)$

$x^{3}+x^{2}-26 x+24$

$=\underline{x^{3}}-x^{2}+\underline{2 x^{2}-2 x}-\underline{24 x+24}$

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

$=x^{2}(x-1)+2 x(x-1)-24(x-1)$

$=(x-1)\left(x^{2}+2 x-24\right)$

$=(x-1)\left(x^{2}+6 x-4 x-24\right)$

[By splitting the middle term]

$=(x-1)[x(x+6)-4(x+6)]$

$=(x-1)(x+6)(x-4)$