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

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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!
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