Factorise $x^{3}-23 x^{2}+142 x-120$

Let $p(x)=x^{3}-23 x^{2}+142 x-120$

We shall now look for all the factors of $-120$ . Some of these are $\pm 1,\,\pm 2,\,\pm 3$ , $\pm 4,\,\pm 5,\,\pm 6$, $\pm 8,\,\pm 10,\,\pm 12$,  $\pm 15,\,\pm 20,\,\pm 24,\,\pm 30,\,\pm 60$

By trial, we find that $p(1)=0 .$ So $x-1$ is a factor of $p(x)$

Now we see that $x^{3}-23 x^{2}+142 x-120=x^{3}-x^{2}-22 x^{2}+22 x+120 x-120$

$=x^{2}(x-1)-22 x(x-1)+120(x-1)$         (Why ?)

$=(x-1)\left(x^{2}-22 x+120\right) $                      [Taking $(x-1) $  common]

We could have also got this by dividing $p(x)$ by $x-1$

Now $x^{2}-22 x+120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have :

$x^{2}-22 x+120 =x^{2}-12 x-10 x+120$ 

$=x(x-12)-10(x-12)$

$ =(x-12)(x-10)$

So,          $x^{3}-23 x^{2}-142 x-120=(x-1)(x-10)(x-12)$