Решить методом интервалов (x-4)(8-x)(x+5) больше 0

[latex](x-4)(8-x)(x+5)\ \textgreater \ 0, \\ -(x-4)(x-8)(x+5)\ \textgreater \ 0, \\ (x-4)(x-8)(x+5)\ \textless \ 0, \\ (x-4)(x-8)(x+5)=0, \\ \left[\begin{array}{c}x-4=0,\\x-8=0,\\x+5=0;\end{array}\right. \left[\begin{array}{c}x=4,\\x=8,\\x=-5;\end{array}\right. \\ \begin{array}{c|ccccccc}x&(-\infty;-5)&-5&(-5;4)&4&(4;8)&8&(8;+\infty)\\(x-4)(8-x)(x+5)&-&0&+&0&-&0&+\end{array} \\ x\in(-\infty;-5)\cup(4;8).[/latex]