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Write the solution set of each inequality if x is an element of the set of integers.$x^{2}-x-6<0$

$\{x |-2<x<3\},\{-1,0,1,2\}$

Algebra

Chapter 1

THE INTEGERS

Section 8

Quadratic Inequalities

The Integers

Equations and Inequalities

Polynomials

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in this exercise, we want to write the solution set of this inequality. If X is an element in the set of injures, so first thing we want to do is just factor Expert minus X minus six. The easiest way to do this is look at the factors of six. So already we can see we have six and one. We have two and three. And you think six. Someone probably won't give us our answer here. Because if we multiply six and won and we think about it in terms of factoring, we're gonna either gets have been like five X or my seven X. It's not gonna work using three and two. We're probably going to end up with this negative X term so we can have X minus three times explodes, too, because if we well, by that out we get X squared plus two x minus three x, which gives us this negative X minus six. This is less than zero. Let me see that are critical. Values are at X equals three and X equals negative to Celeste drawer number one drawing a number line. We have three and negative too. So first we want to pick some values above three and then in between negative two and three and then below negative to plug them in and see what we get. So if we plug in a number above three, like four of them and we get four minus three, which is a positive number and four plus two, which is a positive number, positive times positive gives you a positive value. And of course, that is not less than zero. So we say that this region does not work. Now, if we pick something in between negative two and three like zero, we plug that and we get a native number times a positive, which just give us less than zero. So this region right here does work? No. If we pick something below negative to, like negative three, if you plug that in, we get a negative times negative, which will give us a positive value. So this region does not work. So, looking at a number line, we can see that the only reason that does work is when X is between negative, too, and three. And that's your answer.

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