Question 594467
5x(x+1)(x-1)>0

 first, solve for 5x(x+1)(x-1)=0
                       5x=0             x = 0
5x(x+1)(x-1) = 0  -->  x+1 =0     -->   x = -1
                       x-1 = 0          x = 1


now,  we have to test the function  f(x)= 5x(x+1)(x-1) with its roots (  0, -1, 1), by taking any numbers in all those intervals , plugging that in the function and putting the sign of the answer in the interval you took the number from.         ( consider oo as infinity and -oo as minus infinity)

  let's take, -2 from  (-oo,  -1)   f(-2) =  5(-2)(-2+1)(-2-1) = -30
              -1/2 from ( -1, 0)  f(-1/2) = 5(-1/2)(-1/2-1)(-1/2+1)= + 15/8
               1/2 from (0,  1)    f(1/2) =  5(1/2)(1/2-1)(1/2+1) =  -15/8
               2 from ( 1, oo)       f(2) = 5(2)(2-1)(2+1) = + 30
      
       
    we get this   (-oo) ----- (-1) ++++ (0) ----(1) ++++ (oo)         

looking at this , there are only two intervals satisfying that question. those intervals are (-1, 0) and (1, oo)

so the answer is (-1, 0)U(1, oo)