## College Algebra (10th Edition)

The solution set is $\left\{-4, 4\right\}$.
Factor out the GCF (which is 3) on the left side to obtain: $3(x^2-16)=0 \\3(x^2-4^2)=0$ Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain: $3(x-4)(x+4)=0$ Use the Zero-Product Property by equating each factor with a variable to zero to obtain: $\begin{array}{ccc} &x-4=0 &\text{ or } &x+4=0 \\&x=4 &\text{ or } &x=-4 \end{array}$ Therefore the solution set is $\left\{-4, 4\right\}$.