 Today I want to go over how we can solve polynomial inequalities. The first example we are going to look at is x minus one cubed times x plus two squared times x minus three is less than or equal to zero. So the first thing that you want to do whenever solving a polynomial inequality is set it equal to zero. This makes it easy for us because then we just are trying to solve when it's equal to zero rather than a different number. So here that's nice because when it's equal to zero we can just take each factor and set it equal to zero and when we do that we can find the zeros or the x intercepts of the function pretty easily which here will be at x equals one, x equals negative two and x equals three. So what I usually start out doing is putting these values on a number line and I know that on my graph these are going to be x intercepts. Now you can pause the video and type the equation into your graphing calculator if you want but I'm going to talk through how we can think about what the graph will look like just based on the equation to get just a rough sketch of the graph. So if we look here at all of the x values notice that all of them have a leading coefficient of one and so what that means for us is that means the leading coefficient of our graph will be one and so it has a positive leading coefficient and it will end in the positive direction. And what we can look at is we can look at the degree on each factor. Notice here at the x minus three this has a degree of one or it's just a single zero and so when it's a single zero it goes straight through the graph. Then if we look at our zero at one the x minus one notice the degree of that factor is third power and so with our third power what happens with the graph is it goes through and it just kind of does its flattening out but it still goes through. So we would get some curve at the bottom then it would kind of curve flat and going through at one and it would probably go up like this then we know it has to come back down around negative two and if you look right at negative two or x plus two at our factor the degree on this factor is a squared and anytime it's an even degree it will have a turning point right at the zero. And so our graph we know will go up and turn right through there at negative two. So this is just a really rough sketch of what the graph would look like but we can use this sketch in order to help us figure out when the graph is less than or equal to zero. So less than or equal to zero is going to be anywhere where the graph is equal to or below the x-axis or below our number line here. So what I'm going to do as I go through I'm just going to go along the x-axis and when I get to a point where it hits the x-axis or it's below then I'm going to shade it in with this highlighter. So notice here at one it drops below until we get to three and then nothing else on the graph is below. So we just have a few things that are highlighted and that will help give us our solution. So notice right at negative two it includes just the negative two because it's equal to zero. So there's no interval there just the negative two. And if we keep going along notice there at one again it's equal to zero but then all the way from one to three it's below. And I can include zero and that's why I'm using brackets and then as I finish past three it never goes below again and so my solution will just be x equals negative two or x equals a value between one and three. Okay let's look at one more example. Okay here's a second example for us to look at. What we're going to do here is just like before we're going to first set the inequality equal to zero. So I'm going to move the 4x cubed and the 10x to the other side of the equation and I'm just going to write it in standard form. So I have x to the fourth minus 4x cubed minus 3x squared plus 10x and that is supposed to be an eight there at the end so plus eight is greater than zero. So this one's a little bit different from before because it's not in factored form. So if you wanted to try to factor it you could but I think probably the easiest thing to do since we're comparing it to zero is going to be to look at your graph and find any x intercepts that you can on your graph. So pause the video for a moment, type it into your y equals on your graph and find any of the x intercepts from your graph. When I typed it in to my calculator I had x intercepts at negative one, two and four and so what I like to do is label those on a number line and it's just a rough sketch so that's why I'm just putting in those values and then using the graph or you can again look at the equation but sketch just a rough picture of what this graph looks like. So when I look at the graph it starts with a positive end behavior to the left then at negative one it must be a double zero because it just has a turning point right at negative one goes up really high, down through two, down very low and then up again through four. So that's just a rough sketch of what our graph will look like but using this graph we can now compare and find out when it is greater than zero. So this time with this example we want it to just be greater than which means anywhere where it's above the x axis so I'm going to start tracing here because it's already greater than zero so I'll highlight and then notice I'm not going to include negative one because we just want greater than so I'm breaking but then it is greater than again until we get to two and it falls below and then at four right after four again we get a section of the graph that's greater than so we have the highlighted portion here at the top and from that we can solve and write our intervals. So x is going to be greater than zero starting at negative infinity so we're reading from left to right so negative infinity is over here and then we go all the way until we get to negative one but we don't include negative one. There's just a break there because then again at negative one it's greater than until we get to two but again we don't include two then it's below the x axis until we get to four but at four it's not included so we do a parenthesis and then the rest of the graph will be above the x axis so it will be from four to infinity. So three different intervals is what should be included in the solution for this problem.