$\left(x-\frac{1}{2}\right)\left(x+4\right)\left(x^2-9\right)=0$
$\begin{array}{ccccc}x-\frac{1}{2}=0& \vee & x+4=0& \vee & x^2-9=0\\ x=\frac{1}{2}& & x=-4& & x^2=9\\ & & & & x=3\vee x=-3\end{array}$

 

$\left(x-1\right)\left(x+3\right)\left(x^2-16\right)=0$

$\begin{array}{ccccc}x-1=0& \vee & x+3=0& \vee & x^2-16=0\\ x=1& & x=-3& & x^2=16\\ & & & & x=4\vee x=-4\end{array}$