## Elementary Algebra

The solution set is $\left\{-4, 0, 4\right\}$.
Factor out $3x$ to obtain: $3x(x^2-16)=0 \\3x(x^2-4^2)=0 \\3x(x-4)(x+4)=0$ RECALL: A difference of two squares can be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares to obtain: $3x(x-4)(x+4)=0$ Equate each factor to 0, and then solve each equation to obtain: $3x=0 \text{ or } x-4=0 \text{ or } x+4 = 0\\\ x=0 \text{ or } x=4 \text{ or } x=-4$ The solution set is $\left\{-4, 0, 4\right\}$.