Answer::
[tex](-\infty,-4)\cup(7,\infty)[/tex]Explanation:
Given the function:
[tex]3\left(x+4\right)\left(x-7\right)^3[/tex]We want to determine the interval in which the function is greater than 0.
First, solve for the critical values.
[tex]\begin{gathered} 3\left(x+4\right)\left(x-7\right)^3=0 \\ x+4=0,x-7=0 \\ x=-4,x=7 \end{gathered}[/tex]So, the following are the possible intervals:
[tex]\begin{gathered} (-\infty,-4) \\ (-4,7) \\ (7,\infty) \end{gathered}[/tex]We test each of the intervals:
[tex]\begin{gathered} (-\infty,-4),\text{ At }x=-5:3\left(x+4\right)\left(x-7\right)^3=5184>0 \\ (-4,7),\text{ At x}=0:3\left(x+4\right)\left(x-7\right)^3=-4116<0 \\ (7,\infty),\text{ At }x=8:3\left(x+4\right)\left(x-7\right)^3=36>0 \end{gathered}[/tex]Therefore, the function is greater than zero in the intervals:
[tex](-\infty,-4)\cup(7,\infty)[/tex]The result can be confirmed using the graph below:
