Solve the inequality and write down the solution in interval notation x (x-3) (x+1) <0

Since it's so nicely grouped, we can work with it! For the equation to equal 0, x=0, 3, or - 1 (since x-3 and x+1 equal 0 when plugged in with 3 and - 1 respectively). All we have to do is plug in numbers before, between, and after these numbers and apply it to the rest of them. Since - 1 is the smallest number of the group, we can plug in a number below that (for this example, - 5) and plug it in to get - 8*-5*-4 = something negative since it contains an odd number of negative numbers. Therefore, anything less than 1 is negative. For between - 1 and 0, we get x=-0.5 equals - 0.5*-3.5*0.5=something positive (since it has an even amount of negative numbers), proving that everything between - 1 and 0 here is positive. For something between 0 and 3, we can plug 1 in to get 1*-2*2 = something negative. Do you see a pattern here? It's negative, then positive, etc ... Therefore, if the number is greater than 3 it is positive. Reviewing a bit, we can see that (-inf, - 1) is negative as well as (0,3), making the interval notation (-inf, - 1) U (0, 3) since when you plug - 1, 0, and 3 in it is 0, not less than 0!