please help having trouble with this

(x+7)(x-12)(x+5)>0

the solution set is or is it all real numbers.

All parts of the uneqiations must be large of 0

Less part is ( x - 12 )

Solution x > 12

multiply and simplify by factoring assume all expression under radicals represent non negative numbers

¡¼∛y¡½^7 ∛(¡¼81y¡½^8 )

knowing the general shape of a cubic, and knowing that the intercepts are

x = -7, 12 , and -5

we can see that
(x+7)(x-12)(x+5)>0

has solution -7 < x < -5 OR x > 12

To solve the inequality (x+7)(x-12)(x+5)>0, we need to find the values of x that make the expression greater than zero.

We can start by noting that the product of three factors is greater than zero only if either all three factors are positive or one factor is positive and the other two factors are negative.

1. Setting each factor greater than zero:
(x+7)>0 => x>-7
(x-12)>0 => x>12
(x+5)>0 => x>-5

By setting each factor greater than zero, we can see that x needs to be greater than -7, 12, and -5 respectively.

2. Setting two factors negative and the third factor positive:
(x+7)<0 => x<-7
(x-12)<0 => x<12
(x+5)>0 => x>-5

By setting two factors negative and the third factor positive, we can see that x needs to be less than -7 and between -5 and 12.

Combining both cases, we have the solution:

x<-12, -7<x<-5, x>12

So, the solution set to the inequality (x+7)(x-12)(x+5)>0 is x<-12 or -7 < x < -5 or x > 12. In interval notation, it can be written as (-∞, -12) U (-7, -5) U (12, ∞).

Therefore, the solution set is all real numbers except for the interval between -12 and -7, and the interval between -5 and 12.