Question:
Obtain all zeros of $1(x)=x^{3}+13 x^{2}+32 x+20$, if one of its zeros is $-2$.
Solution:
Since $-2$ is one zero of $f(x)$.
Therefore, we know that, if $x=\alpha$ is a zero of a polynomial, then $(x-\alpha)$ is a factor of $f(x)=x+2$ is a factor of $f(x)$.
Now, we divide $f(x)=x^{3}+13 x^{2}+32 x+20$ by $g(x)=(x+2)$ to find the others zeros of $f(x)$.
By using that division algorithm we have, $f(x)=g(x) \times q(x)+r(x)$
$x^{3}+13 x^{2}+32 x+20=(x+2)\left(x^{2}+11 x+10\right)+0$
$x^{3}+13 x^{2}+32 x+20=(x+2)\left(x^{2}+10 x+1 x+10\right)$
$x^{3}+13 x^{2}+32 x+20=(x+2)[x(x+10)+1(x+10)]$
$x^{3}+13 x^{2}+32 x+20=(x+2)(x+1)(x+10)$
