Answer
{$-12,-4,0$}
Work Step by Step
First, we factor out $x$ from the expression as it is common to all the terms:
$x^{3}+16x^{2}+48x=0$
$x(x^{2}+16x+48)=0$
Using the rules of factoring trinomials to factor the polynomial, we obtain:
$x(x^{2}+16x+48)=0$
$x(x^{2}+4x+12x+48)=0$
$x[x(x+4)+12(x+4)]=0$
$x(x+4)(x+12)=0$
Now, we equate the factors to zero:
$x(x+4)(x+12)=0$
$x=0$ or $(x+4)=0$ or $(x+12)=0$
$x=0$ or $x=-4$ or $x=-12$
Therefore, the solution set is {$-12,-4,0$}.
