Factorise: $x^{3}+13x^{2}+32x+20$

(D) Let $p(x) = x^{3}+13x^{2}+32x+20$.
By trial,let us find $p(1)$:
$p(1) = (1)^{3}+13(1)^{2}+32(1)+20 = 1+13+32+20 = 66 \neq 0$.
Now,let us find $p(-1)$:
$p(-1) = (-1)^{3}+13(-1)^{2}+32(-1)+20 = -1+13-32+20 = 0$.
$\therefore$ By the factor theorem,$(x+1)$ is a factor of $p(x)$.
Now,divide $p(x)$ by $(x+1)$:
$x^{3}+13x^{2}+32x+20 = (x+1)(x^{2}+12x+20)$.
Factorize the quadratic expression $x^{2}+12x+20$ by splitting the middle term:
$x^{2}+12x+20 = x^{2}+2x+10x+20$
$= x(x+2)+10(x+2)$
$= (x+2)(x+10)$.
Therefore,the factors are $(x+1)(x+2)(x+10)$.