## Elementary Algebra

{$-12,-4,0$}
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 {$-12,-4,0$}.