Given the inequality:
[tex](x-7)^2(x+5)>0[/tex]
This is:
[tex]\begin{gathered} (x-7)^2>0 \\ \text{and } \\ x+5>0 \end{gathered}[/tex]
Then solve for x in each one:
[tex]\begin{gathered} (x-7)^2>0 \\ \sqrt[]{(x-7)^2}>\sqrt[]{0} \\ x-7>0 \\ x-7+7>0+7 \\ x>7 \end{gathered}[/tex]
And
[tex]\begin{gathered} x+5>0 \\ x+5-5>0-5 \\ x>-5 \end{gathered}[/tex]
So, identify the intervals:
[tex]-57}[/tex]
In interval notation, the answer is:
[tex](-5,7)\cup(7,\infty)[/tex]
