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

(N/A) Let $p(x) = x^{3}-23 x^{2}+142 x-120$.
We look for factors of the constant term $-120$. 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 $p(1) = 1^{3} - 23(1)^{2} + 142(1) - 120 = 1 - 23 + 142 - 120 = 0$. Therefore,$(x-1)$ is a factor of $p(x)$.
Now,we rewrite the polynomial:
$x^{3}-23 x^{2}+142 x-120 = x^{3}-x^{2}-22 x^{2}+22 x+120 x-120$
$= x^{2}(x-1) - 22x(x-1) + 120(x-1)$
$= (x-1)(x^{2}-22x+120)$.
Next,we factorise the quadratic $x^{2}-22x+120$ by splitting the middle term:
$x^{2}-22x+120 = x^{2}-12x-10x+120$
$= x(x-12) - 10(x-12)$
$= (x-12)(x-10)$.
Thus,the complete factorisation is $x^{3}-23 x^{2}+142 x-120 = (x-1)(x-10)(x-12)$.