 This video will talk about solving polynomial and rational inequalities. We want to solve these polynomial inequalities. We've looked at absolute value, we've looked at quadratics. The absolute values, we were allowed to do it symbolically. But when we started with quadratic, then we needed to use it either graphically or do some kind of testing because we're going to have boundary points that will give us these intervals of values that could be true or false. So we want to take this problem and it asks us to factor it and I'm going to factor it using synthetic division. We've been talking about that in this chapter so we might as well use it. I'm looking at negative 12 to help me figure out what to try because I know one or negative one doesn't work. So I want to try negative two and I put my negative two there and I put in all my coefficients and I'm ready to go. So I'm going to bring down the one and multiply and then add and then multiply and I add and I multiply and I add and thankfully I get a zero down here so I know that this is one of my zeros. So I can even put it over here on my graph. And then I can take this and I know that it's remember constant x and x squared. So x squared minus x minus six and I want to factor that. And I could take my quadratic formula and my calculator or this is simple enough because it's just an x squared. So I'm going to take x and x and minus three and plus two that will add up to negative one. So this tells me if I set it equal to zero that I have a zero at positive three and a zero at negative two, which means I have two here. So when I factor this polynomial up here it's actually x minus three x plus two quantity squared because one of the second factor was what I divided by. Now we know that we have these two zeros and we need to figure out what happens between those things and we also need to think about that it's going to be in our case brackets. So that will help me out knowing how to write my final answer. So I have over here from negative infinity to negative two and then I have from negative two over to positive three and I also have from positive three to positive infinity and I want to know what happens in all those regions. So I'm coming over here to my calculator and why one is where this polynomial is and I want to look at something smaller than negative two. So over here it's got a negative value and if I look in between the two and I look at like zero I see that it also has a negative value. Here's my negative, there's my zero and then until I get to the next zero it's still a negative value and then if I look at three I need to look beyond three and I see that those are all positive. Now if I look at it says over here to determine the end behaviors this is a positive cubic graph so it should be a down up, starts down and ends up and I want to talk about the multiplicities. We had a multiplicity of two over here at negative two and that is why over here we started negative because that's the way the graph is supposed to be. But then at that two it had to bounce so it stayed negative and this isn't a perfect graph but then it does something to come through this one and then end up positive. So this is just a guess of my graph but I can see now if I'm looking graphically that sure enough below the x-axis is from negative infinity to three and then above it will be from three to positive infinity. We want greater than zero so that means we want this over here so our answer is going to be x is an element of three including two infinity. And if we come over here and look at the actual graph if we graph it we see that sure enough it's below everywhere until we get to that three and then it's above just like we thought not exactly the same graph we graphed but we could see the shape is about right. Alright so now we want to solve this inequality using this table which is basically what we did but we kind of did a combination of both. So we need to get this to be zero on one side and everything else on the other side so bring over my 5x squared and my 2x and then I have my plus eight less than or equal to zero. I was just adding and subtracting across the equal sign so I didn't have to change the inequality and this one we need to figure out where our zeros are and we could do it a couple ways. We could do synthetic division we could try to see if we could factor it some other way or my y2 here is where I have my zeros and I see that I have a zero at negative one and I have another zero at x equal four and remember I need four of them I mean three of them so I need to go and I missed one. Here's the other one over here at two equals zero or x equal two. So all of those were found pretty quickly I found three I didn't have to worry about multiplicities because I found three so that makes my intervals this is my smallest value at negative one so I'm going to go from negative infinity to negative one and it will be including the negative one because of my inequality and I'm going to go then the next smallest one would be two so I'm going to start at that negative one and go over to two and include it and then I will go from two to four those should be all the values I need so a test value is smaller than negative one I'll try negative two and then I'll go look at my table to figure it out instead of plugging and chugging from negative one to two I can see a zero happens in there so I'm going to try zero and between two and four well that would be three and actually I should do from four to infinity I forgot about that one and so I'll try five. So negative two if I go up one more remember I'm in this y2 column and at negative two I have negative 24 so I have a negative and my inequality says less than or equal to zero so yes this one works and when I've put in zero I go over here and I see that I have a positive eight so that gives me something that I don't want between two and four I tried three and that gives me a negative value so I want that one as well and then if I go down a little bit farther to five then I see that I have a positive again which again I don't want so I want to say that I have from negative infinity to negative one because that one worked and then union I have two to four and we really should say that x is an element of these intervals now we want to look at rational inequalities and we want to use a number line again and behavior at the zeroes and the asymptotes and you have to remember about the asymptotes now because remember vertical asymptotes can't be included so I'm going to factor all these things remember factoring the top and then setting it equal to zero will give me my x intercept so x is equal to negative three is a zero and then I have x minus two so x is going to be equal to two and that one's going to be a vertical asymptote so I'm a number line here I have negative three and I have two my inequality says less than or equal to so that means I can include my zero so it might be, in fact I changed my color here it might be this direction but it might be this direction but you never can include vertical asymptote so I have to have a parenthesis there on either side of it because it will never be two it'll get close, never be there so then we need to look and see what the values are again just like we did with the other one and I'm going to come and move over to y three because I happened to have put this one in my calculator my value is a little quicker and I want to choose like negative four if you think about these these are all going to be degree one so everything's going to change sign on either side of these so that helps us out so at negative four I see that I have a positive value and then I want to choose something between negative three and two so I'll try zero and that gives me a negative four so it's a negative value and then I get the error notice you have an error there because that's my vertical asymptote and then if I go below that I'm going to have a positive again so I try three and it gives me a positive value then we look and see we want it to be less than or equal to zero so that's going to be in here and use what we have on our number line of bracket and a parenthesis negative three goes here two goes here and x is an element of that trying again we're going to factor the top and we have x that would be plus four and minus two so my zeros are going to be at negative four and positive two and if I factor the bottom down here I could do again my synthetic division on it to get it down real quick I factor quicker that way one five three and negative nine six times one is going to be six add that I get nine times one is positive nine and there's my zero so I have a zero at one but it's a vertical asymptote so I'm going to be sure I put my parenthesis in there right away this is going to be x squared plus six x plus nine and if I go to my quadratic formula in here just to do something different I'm going to put in my A as one and B as positive six and C as positive nine and I find out that I have one real zero at negative three which means that this is really x plus three quantity squared so I have x equal negative three and it's not going to change here it's not going to change sign because it's an even exponent so where's my negative three it comes in here somewhere it can't be that one either that'll have parenthesis which says we're different colors these are my vertical asymptotes and blue and I'll put my zeros in red so it could be this way but it could be this way at negative four it could be to the left or to the right a positive two and then I just have to check again for my values to see what's happening we come over here and we look at the table second graph so I have my equation in here and it happens to be in y four and I need to make sure that I can see everything there oops, I need to come over here to the x's and come over to y four at negative five I can test negative five and I've got a negative that's a negative here and that satisfies my inequality so I like that one between negative four and negative three we have the error happening in here but we're going to have to go in we will have to go and check on that one in a moment I'll come back to that one between negative three and one we can look at zero and that's a positive so that one we don't want remember at this zero of negative three we didn't change sign so if it's a positive on this side it'll be a positive on this side so that helps us the one it did change so it should be a negative in here and if we check it actually we can't look between one and two again yet but we can look beyond two and it gives us positives and this was an odd exponent it was just a one just a degree one on that zero so it should be a negative it should change signs from positive to negative and then from negative to positive I like that interval too so we have from negative infinity up to negative four including negative four X is an element of union we have one but notice it's a parenthesis because that was an asymptote so parenthesis one comma two mass problem max is the correct solution with the inequality when g of X is greater than zero if we look at this we have a zero here at negative four we have a zero here that looks like maybe negative one half or something to that effect and then we have one again here at positive four and this negative one half would be point five so negative we want it to be greater so we want it to be between negative four and negative one half so it starts at negative four and it should include it so it's not going to be this one this one starts at negative four inclusive and it goes to negative one half so far so good and then we also want this part which starts at four and goes to infinity and it includes that one so we would say D