Factorise : $x^{3}+13 x^{2}+32 x+20$
$x^{3}+13 x^{2}+32 x+20$
We have $p ( x )= x ^{3}+13 x ^{2}+32 x +20$
By trial, let us find : $p (1)=(1)^{3}+13(1)^{2}+32(1)+20$
$=1+13+32+20=66 \neq 0$
Now $p (-1)=(-1)^{3}+13(-1)^{2}+32(-1)+20$
$=-1+13-32+20=0$
$\therefore$ By factor theorem, $[ x -(-1)],$ i.e. $( x +1)$ is a factor $p ( x )$
$\therefore \quad \frac{x^{3}-13 x^{2}-32 x-20}{(x+1)}=x^{2}+12 x+20$
or $x^{3}+13 x^{2}+32 x+20 =(x+1)\left(x^{2}+12 x+20\right)$
$=(x+1)\left[x^{2}+2 x+10 x+20\right] $
[Splitting the middle term]
$=(x+1)[x(x+2)+10(x+2)]$
$=(x+1)[(x+2)(x+10)]$
$=(x+1)(x+2)(x+10)$
Factorise : $x^{3}-2 x^{2}-x+2$
Find the zero of the polynomial : $p(x) = x + 5$
Use suitable identities to find the products : $(x+8)(x-10)$
If $x+y+z=0,$ show that $x^{3}+y^{3}+z^{3}=3 x y z$.
Factorise : $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$