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Solving a Rational Inequality

Pearson
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Hi, my name is Rebecca Muller. During this session, we're going to look at polynomial and rational inequalities. But to begin, we're going to review what we already know about solving linear inequalities. Here's an example. If we start with 3 minus x is greater than 0, what can we do to solve this linear inequality? Well, one option is to add x to both sides. So we're going to end up adding x to the left-hand side to just leave us with 3. And when we add x to the right-hand side, we end up with x. So this says that 3 is larger than x, which is comparable to saying that x is less than 3. Now, there's an alternative to solving this. We could have said negative x is going to be greater than-- instead of moving the x to the right-hand side, we think about subtracting 3 from both sides. But now to solve for x, we need to divide by a negative 1. And when we divide by negative values on both sides of an inequality, we have to remember to switch the inequality symbol. That is, division by negative 1 is going to leave us with x here. It's going to switch my inequality symbol to a less than instead of a greater than. And it will divide negative 3 by negative 1 to give us 3. So of course we end up with the same result, but the process was a little bit different. Now let's look at what is going to go on when we look at polynomial inequalities. So if I have x squared minus 4 is less than 0, well, I can factor the left-hand side. That's a difference of 2 square. So this is x minus 2 times x plus 2. At this point, were this an equation, we would set each of the factors equal to 0. So if it had an equals sign, that's what we would do. Can we do the same thing here? Can we take each factor and say that it has to be negative? The answer is no. If both factors are negative, then the product is going to be positive. So that means we need to have other options. I mean, we have to think about maybe the first one is positive and the second one is negative, or vice versa. We have all these cases to consider. So we're going to learn a method by which we can keep track of those cases in order to solve this polynomial inequality. So now let's look at our subtopics for this session. We're going to look at creating a sign chart to solve an inequality. So this is going to be our new method for solving polynomial and rational inequalities. We'll then be using a graph to solve an inequality, another method. And then we'll just look at solving some more versions of polynomial inequalities, and then move on to solving rational inequalities. So let's go back to the problem that we had earlier. That is, the x squared minus 4 is less than 0. And we had it factored, x minus 2 times x plus 2 less than 0. And I want to go from this factored form. Now, the process we're going to use right now is to create something called a sign chart. That's just going to be keeping track of the signs of the factors. So the first thing we need to do is find out where those factors would actually be giving us a product that is equal to 0. So if we take x minus 2 times x plus 2 and set it equal to 0, we can see that we get two solutions. The first factor will equal 0 when x equals 2. The second factor will equal 0 when x equals negative 2. Now, follow along for a minute. We'll go through the process, and then talk about a lot of the reasoning after you see what the process is. I'm going to take these two numbers, and I'm going to place them on a number line in the order that they normally would appear. In that, negative 2 is going to be smaller than positive 2, so we want to make sure that when I draw this number line, I'm going to have my negative 2 to the left of the positive 2. Now, I don't really want to put an entire number line down here. I'm not going to put the 0 or whatever else. I'm only going to put the values that are pertinent to the solution that I'm worried about. To the upper part of this number line, I'm going to just draw some dashed lines. What this does is it sections off my number line into three intervals. The numbers to the left would be all values that are smaller than negative 2. The numbers in the middle would be all values between negative 2 and positive 2. And the numbers to the right would be all values that are larger than 2. The next thing I'm going to do is-- give yourself some space here. I'm going to make sure I'm kind of off the chart here. And I'm going to list the factors of that polynomial that we were looking at. So I have my factors x minus 2 and x plus 2. And I'm going to try to figure out what the sign is on each of these factors in each of these intervals so that then I can decide what the sign of the product is. The good thing about signed numbers is we know that if we multiply two numbers together that have the same sign, the product is going to be positive. If we multiply together two numbers that have opposite signs, the product will be negative. So we're going to use that fact. Here's how we'll proceed. We're going to pick a number in each interval. So for instance, if I start here in the interval to the right, all I need do is pick a number that is larger than 2. Let's say we'll pick 3. I'm going to substitute that value into the factor, but the thing that I'm worried about is whether or not the result is going to be positive or negative. So 3 minus 2 equals 1, but what I'm interested in is that 1 is a positive number. 3 plus 2 is going to be 5, but again, what I'm really interested in is the fact that that result is positive. And I'm going to do the same thing for each of the intervals. So between negative 2 and positive 2, I can choose the value 0, which is normally pretty easy to substitute into any kind of expression. 0 minus 2 is going to be negative. 0 plus 2 is positive. Pick a number smaller than negative 2 in order to look at the last interval here to the left, so for instance, negative 3. Negative 3 minus 2 is negative. Negative 3 plus 2 is also negative. At this point, I know what's going on with each of the factors in each of the intervals. I can now look at the sign of the product. When I multiply a positive number times a positive number, the result is positive. When I multiply a negative number times a positive number, the result is negative. And a negative times a negative is going to be positive. Now, let's just talk a minute about why this works. This is a linear factor. It's going to change sign one and only one time. If I choose numbers that are larger than 2, this factor is going to always be positive. If I choose numbers smaller than 2. Those values will always be negative. Look at this second factor, x plus 2. It's equal to 0 at the value of negative 2. Choosing numbers larger than negative 2 gives me positive results. Choosing numbers small than negative 2 gives me negative results. So when we look at our product, we can now see that I'm choosing to look at values where the product is going to be negative. That's going to occur in the middle interval. Those would be all numbers between negative 2 and positive 2. So we can write that down in interval notation from negative 2 to positive 2. And the next question is, do we include those endpoints? The endpoints were the values that made this polynomial equal to 0. Here we have a strict inequality. So we are not going to include those endpoints, and this is going to be the answer to the inequality. When is x squared minus 4 less than 0? Any time I choose a number in the interval from negative 2 to positive 2 exclusive of the endpoints. So now we're going to look at a second method for solving inequalities. We're going to look at the associated graph when we look at an inequality. So if you recall, we had the inequality x squared minus 4 is less than 0. The associated graph is f of x equals x squared minus 4. And you'll recall that's a parabola that's been shifted down 4 units. Now, if we look at that inequality, we notice that now what we're looking at is f of x being less than 0. And remember, our f of x is really our y value. So what we have here is the y value on the graph being negative. That's going to occur in quadrants III and IV. What will that look like if we just look at that section of the graph? We're going to end up having a section of the graph that is below the x-axis. And you can now see from the graphical representation that if we look at values of x that are between negative 2 and positive 2, that whole area of the graph will be under the x-axis and the y value will be negative. So we can say-- the same result as we had previously-- the solution to this inequality will be from negative 2 to positive 2. And again, we do not include the endpoints, because that's where the f of x would equal 0. It's time for a quick quiz. The graph of y equals f of x is depicted below. We have given an xy plane with a graph that begins in quadrant 3 goes through the x-axis at the value of negative 2 moves upward until it ends up going to the high value of positive 1 on the y-axis. It was back down to end up hitting x-axis at 3 dips below the x-axis comes back up hits the x-axis at 5 and then continues upward from that point on. There's a smooth curve and it's continuous and all points. We want to use the graph to solve the inequality f of x is greater than or equal to 0. Is the correct answer. A. We're given an open bracket negative 2, 3 closed bracket union open bracket 5, infinity and then close parentheses. Is the answer B. Open parentheses negative infinity comma 1 with a closed bracket union open bracket 4 comma infinity close parentheses. or C. Open parentheses negative infinity comma negative 2 closed bracket union open bracket 3 comma 5 close bracket. Choose A, B, or C now. You're right, the answer is A. We have an interval to begins at negative 2 ends at 3 including both end points union with an interval that begins at 5 including 5 and goes to positive infinity. Sorry the answer is A, this is the interval that begins at negative 2 goes to positive 3 including both end points union with an interval it begins at 5 and goes to positive infinity including the end point 5. In this problem we're trying to determine where f of x is greater than or equal to 0. This means we're looking for values where we're going to have a y-coordinate greater than or equal to 0. On an xy plane, y coordinates are equal to 0 whenever we have points that or on the x-axis. y-coordinates are greater than 0 if we're looking at points that are in either quadrant 1 or in quadrant 2. So a point considering the graph we note that we have x intercept at negative 2, at positive 3 and it positive 5 those will be part of our solution set. We're also going to note that the graph is above the x-axis as long as I am looking at values that are between negative 2 and positive 3 and then are greater than the value of 5. So I can choose A because those are going to give me the intervals the 2 intervals that combined show me wherever my y-coordinate would end up being greater than or equal to 0. So we now have two methods for solving polynomial inequalities. And we can see that one of them is to set up a sign chart. The other is to draw a graph. It turns out if you're able to sketch that graph pretty quickly and accurately, then that may be your preferable method. But there are a lot of times when that's not going to be the case. Let's look at another example, and you'll see what I mean. If I look at x cubed plus 16x is greater than or equal to 8x squared, then first of all, I'm going to set this in relation to 0 in order to use the sign chart. So I would rewrite this as x cubed minus 8x squared plus 16x greater than or equal to 0. And you'll notice I wrote it in descending powers of x. Well, I don't really know what the graph of y equals x cubed minus 8x squared plus 16x looks like really fast. So instead, I'm going to use the method of using a sign chart. Now, that does require me to start with factoring. So I'm going to concentrate on factoring the left-hand side of this inequality. First thing I notice is there's a common factor of x. Let's go ahead and factor that out, leaving us with x squared minus 8x plus 16 inside parentheses multiplied times, of course, our x. Inside the parentheses now, we can try to factor further. And you may recognize this as one of the special products, or you could use FOIL method if you prefer, to notice that this is x minus 4 quantity squared. Now, mimicking what we did in our other example, I'm going to take this expression and set it equal to 0 in order to find out those values that I want to put on the number line that are going to be my cutoff values to separate intervals. The first factor, x, could equal 0, or the second factor could equal 0. And that's going to occur when x equals 4. So these are the two numbers I'm going to place on the number line in order to get my sign chart. Let's do that next. Let's go ahead and put it here. I have 0 is smaller than 4, so I'll put it over here to the left. I'm going to go ahead and put some dashed lines up here to separate my intervals, three intervals. I want to place the factors separately in each row. So I have an x, and I have an x minus 4 quantity squared. And then I'll be interested in finding out the sign of their product. So here we go. Let's pick a number in each of the three intervals that we see here, and figure out what that sign would be. If I pick a number bigger than 4, like 5, substituting in for x, that's a positive value. 5 minus 4 is 1. Square it. That's also going to be positive. Pick a number between 0 and 4. I'll just choose the number 2. Substituting for x, I end up with a plus sign. 2 minus 4 is negative 2. When squared, it is positive. Pick a number smaller than 0, like negative 1. Substituting for x, that's going to be negative. Negative 1 minus 4 is negative 5. Squaring that gives us a positive 25. So let's now look at the product. We multiply two positives together to end up with a positive result. Two positives together gives us a positive result. A negative and a positive is a negative result. So now, going to the point where we have the inequalities set in relation to 0-- that's right here in our second step-- we notice that what we want to find out is when that expression is going to be positive or equal to 0. So we can see the positive sections here and here. Remember that these values, 0 and 4, were where the expression was actually equal to 0, right here. So we do want to include those values in our solution. So notice if I start at the value 0 and then I continue to 4, and I want to include 4, and then continue further, I'm going all the way to positive infinity. So the solution for the inequality is going to be the interval from 0 to infinity, including 0. Now, what if this had been a strict inequality? What if instead of having greater than or equal to 0, what if it had read x times x minus 4 squared is strictly greater than 0? I would have exactly the same sign chart. So notice that the symbol for the inequality is not going to be what determines what the sign chart looks like. It is going to determine how we read off the answer. The answer to this one cannot include the 0 and cannot include the 4, but I still need these two intervals. So I'm going to write those answers down, those intervals down separately. The first interval starts at 0 and goes to 4, and then I want to combine that with the interval where it starts at 4 and goes to infinity. This would be the solution for the strict inequality, whereas the inequality that includes 0 I could put all together into one inequality symbol, or sorry, one interval symbol. Now, I want to go back to the chart just for a second to point out something. The first factor is linear. Notice that it's going to change signs at 0. So if I pick a number larger than 0, it's positive all the time. If I pick a number smaller than 0, it's going to be negative all the time. The second factor is a quantity that's squared. And of course, we're not including the value where it equals 0, because that's one of our cut-off points. Instead, we just notice that no matter what, if I choose a number and I square it, other than 4, it's going to be a positive result. So another way to fill in the chart is just to notice those things going on with the factors. I could fill in an entire row at a time, instead of having to do it by columns. Now, we've created this sign chart based upon an inequality that had the greater than or equal to symbol. But it turns out that the inequality symbol is not important as far as creating the sign chart. It's only reading off the answers that gives us an important part that we have to consider. For instance, what if this had been a strict inequality, as in x multiplied times x minus 4 quantity squared is greater than 0? Then I would still want to pick up the two intervals where this expression is going to be positive, but I would not want to include the endpoints. That is, I'm going to end up with two intervals. The first interval will begin at 0, and I'll put a comma and then stop at 4. And then I'll have that combined with another interval that begins at 4 and goes all the way to infinity. So the solution to the strict inequality would be the interval from 0 to 4 exclusive union the interval that begins at 4, not including 4, and goes all the way to infinity. So one more thing I want to point out before we take off on another example is if I look at what was happening with the factors, you'll see that the x value, which is linear, is going to change sign one time. It's going to change sign wherever it was equal to 0. So it's going to be positive for all values that are greater than 0, but negative for all values that are less than 0. For the second factor, x minus 4 quantity squared, that's going to be a positive value in all three intervals. Notice we're not including the value of 4. That's one of the cut-offs for the intervals. So I'm going to end up with plus signs going across in my sign chart. So you could actually fill out your sign chart, instead of filling it out in columns, you could fill it out in rows, taking advantage of what you know about those individual factors. Next we're going to look at a rational inequality. Here's our example. We have x squared plus 4x plus 16 in the numerator divided by x plus 5 in the denominator. And that whole rational expression needs to be greater than or equal to 4. Now, there's a couple of things to point out. First of all, notice that we cannot allow our x value to equal negative 5. So we always want to take note of any kind of restrictions we're going to have. So x cannot equal negative 5. Next, we have to think about the fact that if we're doing a sign chart, it always relies on these expressions being set in relation to 0. So our first step for this rational inequality will be to subtract 4 from both sides of the inequality. So we're going to have x squared plus 4x plus 16, divided by x plus 5, minus the value 4. And that's going to be greater than or equal to 0. Well, one more thing to consider is my sign chart relies on having factorization. That is, we're looking at the rules for signed numbers, where they're being multiplied, and in this case, divided. So I cannot have the subtraction of 4. I need to rewrite this expression as a single rational expression. So that means I'm going to have to get a common denominator, which is going to be my x plus 5. And what I'll do in the next step is rewrite it as x squared plus 4x plus 16 divided by x plus 5 minus-- that value of 4 I want to think of as being 4 over 1. And then we can multiply both numerator and denominator by x plus 5. I'll go ahead and put it in parentheses here. And that is now going to be greater than or equal to 0. Let's go ahead and simplify. And as we're doing that, notice my denominator will be the common denominator of x plus 5. So in my numerator, I'll have x squared plus 4x plus 16 minus 4 times the quantity x plus 5. That's being divided by now my common denominator of x plus 5. And that whole expression is greater than or equal to 0. Let's go another step. We need to distribute the 4 to give us in the numerator x squared plus 4x plus 16, now distributing the multiplication by negative 4. We have negative 4x and then minus 20. That whole thing is being divided by x plus 5. And that whole thing has to be greater than or equal to 0. Notice now that we can simplify. In the numerator, we have 4x minus 4x, which we'll end up subtracting out. We also have 16 minus 20, which will end up giving us negative 4. So the numerator condenses to x squared minus 4 divided by in the denominator x plus 5 greater than or equal to 0. I'm going to factor that numerator. We're going to have x minus 2 times x plus 2, that whole thing divided by x plus 5, greater than or equal to 0. Now, the next thing we have to do is think about in our sign chart that we're interested in where those factors are going to be changing signs. And I'm going to think about the numerator as having two factors, x minus 2, x plus 2. So that would give me values of positive 2 and negative 2 as places where those factors would change sign. In the denominator, we notice that that factor will change sign where I have x equals negative 5. So we're going to take those values, and those will be the ones that we will transfer over to a number line. Now here's an inequality for you to try on your own. The inequality is 4 divided by x minus 1 is less than or equal to 3. Pause the video and try this, and then restart it, and we'll see how you did. OK, great. You're back. Let's see what you were able to do with this. The first thing to notice is that it's not set in relation to 0. So we want to begin by subtracting 3 from both sides of the inequality. 4 divided by x minus 1 minus 3 is less than or equal 0. Now, one thing to point out is you cannot choose to multiply both sides of this inequality by the denominator to start with, because when we use multiplication in an inequality, we have to know whether or not the expression is positive or negative. So because we have a variable, we don't know whether that expression is going to be a positive value, in which case you would leave the inequality symbol as is, or whether you would be multiplying by a negative, in which case you would have to switch the inequality. That's why we have to do the subtraction of 3. We now need to get a common denominator, which will be x minus 1. So this will be 4 divided by x minus 1 minus-- we'll write it as 3 times-- so we can think of this as 3 over 1 times x minus 1 over x minus 1, less than or equal to 0. Now I have a common denominator. Let's go ahead and combine our numerators. 4 minus 3 times x minus 1 divided by x minus 1 is going to be less than or equal to 0. And now we'll go ahead and simplify the numerator. Before I continue on, I want to point out a mistake that I've seen before with students, and I want to make sure you're not making that mistake. Don't think that you can cancel the x minus 1's here. This is not a factor of the numerator. And in fact, what you'd be doing is like undoing what you had previously, only in an incorrect manner. You can't just end up with 4 minus 3. You need to recognize that you have to simplify the numerator first. Again, x minus 1 is not a factor of the numerator, and that's why it cannot be divided out. So simplifying gives us 4 minus-- we'll have 3x plus 3 divided by, in the denominator, x minus 1, and that is less than or equal to 0. Simplifying, 4 plus 3 gives us 7. 7 minus 3x is in the numerator. And x minus 1 is in the denominator. That fraction needs to be less than or equal to 0. As we did in previous examples, we want to set both the numerator equal to 0 and the denominator equal to 0 to find out where those particular factors would end up changing sign possibly. So my first 7 minus 3x set equal to 0 gives us 7 equals 3x. Dividing by 3, we get 7/3 equals x. The denominator set equal to 0 gives us x equals 1. So I need to place these two numbers on the number line to create my sign chart. Now, which one is smaller? Well, x equal to 1 is going to be the smaller value, because 1 is the same as 3/3. And 3/3 is certainly smaller than 7/3. So over to my sign chart, I'm going to place the 1 to the left of the 7/3. Set up my intervals, and put my values, or my numerator. That's 7 minus 3x and x minus 1 from the denominator. We're now going to look at that quotient once we figure out the signs of each of these. Let's choose a value larger than 7/3, so for instance, 9/3, which is going to equal the value 3. 7 minus 3 times 3 is going to be 7 minus 9, which is negative. 3 minus 1 is going to be positive. Choosing a value between 1 and 7/3. Well, let's remember that 1 is equal to 3/3. So if you just move up by thirds, you have 3/3, 4/3, 5/3, 6/3 is smaller than 7/3, and 6/3 can be reduced or rewritten as the number 2. Makes it easy to work with. So 7 minus 3 times 2 is going to be 7 minus 6, which is going to be positive. The value 2 minus 1 is still going to be positive. And finally, choose a number smaller than 1 for the interval to the left. That's easy to use the value 0. 7 minus 3 times 0 is positive. 0 minus 1 is negative. And now we can look at what happens if we have a negative number divided by a positive number. The result is negative. A positive number divided by a positive number would be positive. And a positive number divided by a negative number is negative. So we're able to fill in our sign chart using the signs of the numerator and denominator. In this example, we're interested in where that fraction is less than or equal to 0, which means we're interested in the negative intervals. This interval begins from the left at negative infinity, and continues until I get to the number 1. Now, can I include one in my solution set? The answer is no. If we look back at the outset of the problem, we notice that x minus 1 is coming from the denominator. And so therefore, we do not want to ever include that value in our solution set. And then we need this second interval, which is going to begin at 7/3. Now, 7/3 is a value that comes from the numerator. We want to allow the fraction to equal 0. And therefore, 7/3 makes the numerator 0. And that's part of our solution set. So we're going to use a bracket at the 7/3, and then continue that interval to infinity. So the solution set to this problem is negative infinity, 1, with parentheses union bracket 7/3, infinity. I hope you did well on the problem, and it's time now for you to try more on your own. Good luck. It's time for another quick quiz. What is the solution to the inequality given by 5 divided by the quantity x squared plus 1 is less than 0. Is the correct answer? A. Parentheses negative 1 comma 1 close parentheses. B. Open bracket negative 1 comma 1 closed bracket. or C. The null set. Choose A, B, or C. You're right, the answer is C, the null set. Sorry the answer is C, the null set. Let's consider the fraction where we have the value of 5 in the numerator and the denominator x squared plus 1. We want that fraction to be less than 0 which means we want the fraction to be negative. Right now it's because the numerator is a constant value of 5 we know that the numerator is a positive value. In order for the entire fraction to be negative that would require our denominator to be negative so we would have a positive number divided by a negative number. Well this means that x squared plus 1 would have to be less than 0 which means that x squared would have to be a value less than negative 1. For real number values we know that when we square a number we're going to end up either with a positive value or 0 so there's not going to be a real number value where squared is going to be less than a negative 1. Hence the solution here is that there is no solution, the null set is the correct answer.
Hi, my name is Rebecca Muller. During this session, we're going to look at polynomial and rational inequalities. But to begin, we're going to review what we already know about solving linear inequalities. Here's an example. If we start with 3 minus x is greater than 0, what can we do to solve this linear inequality? Well, one option is to add x to both sides. So we're going to end up adding x to the left-hand side to just leave us with 3. And when we add x to the right-hand side, we end up with x. So this says that 3 is larger than x, which is comparable to saying that x is less than 3. Now, there's an alternative to solving this. We could have said negative x is going to be greater than-- instead of moving the x to the right-hand side, we think about subtracting 3 from both sides. But now to solve for x, we need to divide by a negative 1. And when we divide by negative values on both sides of an inequality, we have to remember to switch the inequality symbol. That is, division by negative 1 is going to leave us with x here. It's going to switch my inequality symbol to a less than instead of a greater than. And it will divide negative 3 by negative 1 to give us 3. So of course we end up with the same result, but the process was a little bit different. Now let's look at what is going to go on when we look at polynomial inequalities. So if I have x squared minus 4 is less than 0, well, I can factor the left-hand side. That's a difference of 2 square. So this is x minus 2 times x plus 2. At this point, were this an equation, we would set each of the factors equal to 0. So if it had an equals sign, that's what we would do. Can we do the same thing here? Can we take each factor and say that it has to be negative? The answer is no. If both factors are negative, then the product is going to be positive. So that means we need to have other options. I mean, we have to think about maybe the first one is positive and the second one is negative, or vice versa. We have all these cases to consider. So we're going to learn a method by which we can keep track of those cases in order to solve this polynomial inequality. So now let's look at our subtopics for this session. We're going to look at creating a sign chart to solve an inequality. So this is going to be our new method for solving polynomial and rational inequalities. We'll then be using a graph to solve an inequality, another method. And then we'll just look at solving some more versions of polynomial inequalities, and then move on to solving rational inequalities. So let's go back to the problem that we had earlier. That is, the x squared minus 4 is less than 0. And we had it factored, x minus 2 times x plus 2 less than 0. And I want to go from this factored form. Now, the process we're going to use right now is to create something called a sign chart. That's just going to be keeping track of the signs of the factors. So the first thing we need to do is find out where those factors would actually be giving us a product that is equal to 0. So if we take x minus 2 times x plus 2 and set it equal to 0, we can see that we get two solutions. The first factor will equal 0 when x equals 2. The second factor will equal 0 when x equals negative 2. Now, follow along for a minute. We'll go through the process, and then talk about a lot of the reasoning after you see what the process is. I'm going to take these two numbers, and I'm going to place them on a number line in the order that they normally would appear. In that, negative 2 is going to be smaller than positive 2, so we want to make sure that when I draw this number line, I'm going to have my negative 2 to the left of the positive 2. Now, I don't really want to put an entire number line down here. I'm not going to put the 0 or whatever else. I'm only going to put the values that are pertinent to the solution that I'm worried about. To the upper part of this number line, I'm going to just draw some dashed lines. What this does is it sections off my number line into three intervals. The numbers to the left would be all values that are smaller than negative 2. The numbers in the middle would be all values between negative 2 and positive 2. And the numbers to the right would be all values that are larger than 2. The next thing I'm going to do is-- give yourself some space here. I'm going to make sure I'm kind of off the chart here. And I'm going to list the factors of that polynomial that we were looking at. So I have my factors x minus 2 and x plus 2. And I'm going to try to figure out what the sign is on each of these factors in each of these intervals so that then I can decide what the sign of the product is. The good thing about signed numbers is we know that if we multiply two numbers together that have the same sign, the product is going to be positive. If we multiply together two numbers that have opposite signs, the product will be negative. So we're going to use that fact. Here's how we'll proceed. We're going to pick a number in each interval. So for instance, if I start here in the interval to the right, all I need do is pick a number that is larger than 2. Let's say we'll pick 3. I'm going to substitute that value into the factor, but the thing that I'm worried about is whether or not the result is going to be positive or negative. So 3 minus 2 equals 1, but what I'm interested in is that 1 is a positive number. 3 plus 2 is going to be 5, but again, what I'm really interested in is the fact that that result is positive. And I'm going to do the same thing for each of the intervals. So between negative 2 and positive 2, I can choose the value 0, which is normally pretty easy to substitute into any kind of expression. 0 minus 2 is going to be negative. 0 plus 2 is positive. Pick a number smaller than negative 2 in order to look at the last interval here to the left, so for instance, negative 3. Negative 3 minus 2 is negative. Negative 3 plus 2 is also negative. At this point, I know what's going on with each of the factors in each of the intervals. I can now look at the sign of the product. When I multiply a positive number times a positive number, the result is positive. When I multiply a negative number times a positive number, the result is negative. And a negative times a negative is going to be positive. Now, let's just talk a minute about why this works. This is a linear factor. It's going to change sign one and only one time. If I choose numbers that are larger than 2, this factor is going to always be positive. If I choose numbers smaller than 2. Those values will always be negative. Look at this second factor, x plus 2. It's equal to 0 at the value of negative 2. Choosing numbers larger than negative 2 gives me positive results. Choosing numbers small than negative 2 gives me negative results. So when we look at our product, we can now see that I'm choosing to look at values where the product is going to be negative. That's going to occur in the middle interval. Those would be all numbers between negative 2 and positive 2. So we can write that down in interval notation from negative 2 to positive 2. And the next question is, do we include those endpoints? The endpoints were the values that made this polynomial equal to 0. Here we have a strict inequality. So we are not going to include those endpoints, and this is going to be the answer to the inequality. When is x squared minus 4 less than 0? Any time I choose a number in the interval from negative 2 to positive 2 exclusive of the endpoints. So now we're going to look at a second method for solving inequalities. We're going to look at the associated graph when we look at an inequality. So if you recall, we had the inequality x squared minus 4 is less than 0. The associated graph is f of x equals x squared minus 4. And you'll recall that's a parabola that's been shifted down 4 units. Now, if we look at that inequality, we notice that now what we're looking at is f of x being less than 0. And remember, our f of x is really our y value. So what we have here is the y value on the graph being negative. That's going to occur in quadrants III and IV. What will that look like if we just look at that section of the graph? We're going to end up having a section of the graph that is below the x-axis. And you can now see from the graphical representation that if we look at values of x that are between negative 2 and positive 2, that whole area of the graph will be under the x-axis and the y value will be negative. So we can say-- the same result as we had previously-- the solution to this inequality will be from negative 2 to positive 2. And again, we do not include the endpoints, because that's where the f of x would equal 0. It's time for a quick quiz. The graph of y equals f of x is depicted below. We have given an xy plane with a graph that begins in quadrant 3 goes through the x-axis at the value of negative 2 moves upward until it ends up going to the high value of positive 1 on the y-axis. It was back down to end up hitting x-axis at 3 dips below the x-axis comes back up hits the x-axis at 5 and then continues upward from that point on. There's a smooth curve and it's continuous and all points. We want to use the graph to solve the inequality f of x is greater than or equal to 0. Is the correct answer. A. We're given an open bracket negative 2, 3 closed bracket union open bracket 5, infinity and then close parentheses. Is the answer B. Open parentheses negative infinity comma 1 with a closed bracket union open bracket 4 comma infinity close parentheses. or C. Open parentheses negative infinity comma negative 2 closed bracket union open bracket 3 comma 5 close bracket. Choose A, B, or C now. You're right, the answer is A. We have an interval to begins at negative 2 ends at 3 including both end points union with an interval that begins at 5 including 5 and goes to positive infinity. Sorry the answer is A, this is the interval that begins at negative 2 goes to positive 3 including both end points union with an interval it begins at 5 and goes to positive infinity including the end point 5. In this problem we're trying to determine where f of x is greater than or equal to 0. This means we're looking for values where we're going to have a y-coordinate greater than or equal to 0. On an xy plane, y coordinates are equal to 0 whenever we have points that or on the x-axis. y-coordinates are greater than 0 if we're looking at points that are in either quadrant 1 or in quadrant 2. So a point considering the graph we note that we have x intercept at negative 2, at positive 3 and it positive 5 those will be part of our solution set. We're also going to note that the graph is above the x-axis as long as I am looking at values that are between negative 2 and positive 3 and then are greater than the value of 5. So I can choose A because those are going to give me the intervals the 2 intervals that combined show me wherever my y-coordinate would end up being greater than or equal to 0. So we now have two methods for solving polynomial inequalities. And we can see that one of them is to set up a sign chart. The other is to draw a graph. It turns out if you're able to sketch that graph pretty quickly and accurately, then that may be your preferable method. But there are a lot of times when that's not going to be the case. Let's look at another example, and you'll see what I mean. If I look at x cubed plus 16x is greater than or equal to 8x squared, then first of all, I'm going to set this in relation to 0 in order to use the sign chart. So I would rewrite this as x cubed minus 8x squared plus 16x greater than or equal to 0. And you'll notice I wrote it in descending powers of x. Well, I don't really know what the graph of y equals x cubed minus 8x squared plus 16x looks like really fast. So instead, I'm going to use the method of using a sign chart. Now, that does require me to start with factoring. So I'm going to concentrate on factoring the left-hand side of this inequality. First thing I notice is there's a common factor of x. Let's go ahead and factor that out, leaving us with x squared minus 8x plus 16 inside parentheses multiplied times, of course, our x. Inside the parentheses now, we can try to factor further. And you may recognize this as one of the special products, or you could use FOIL method if you prefer, to notice that this is x minus 4 quantity squared. Now, mimicking what we did in our other example, I'm going to take this expression and set it equal to 0 in order to find out those values that I want to put on the number line that are going to be my cutoff values to separate intervals. The first factor, x, could equal 0, or the second factor could equal 0. And that's going to occur when x equals 4. So these are the two numbers I'm going to place on the number line in order to get my sign chart. Let's do that next. Let's go ahead and put it here. I have 0 is smaller than 4, so I'll put it over here to the left. I'm going to go ahead and put some dashed lines up here to separate my intervals, three intervals. I want to place the factors separately in each row. So I have an x, and I have an x minus 4 quantity squared. And then I'll be interested in finding out the sign of their product. So here we go. Let's pick a number in each of the three intervals that we see here, and figure out what that sign would be. If I pick a number bigger than 4, like 5, substituting in for x, that's a positive value. 5 minus 4 is 1. Square it. That's also going to be positive. Pick a number between 0 and 4. I'll just choose the number 2. Substituting for x, I end up with a plus sign. 2 minus 4 is negative 2. When squared, it is positive. Pick a number smaller than 0, like negative 1. Substituting for x, that's going to be negative. Negative 1 minus 4 is negative 5. Squaring that gives us a positive 25. So let's now look at the product. We multiply two positives together to end up with a positive result. Two positives together gives us a positive result. A negative and a positive is a negative result. So now, going to the point where we have the inequalities set in relation to 0-- that's right here in our second step-- we notice that what we want to find out is when that expression is going to be positive or equal to 0. So we can see the positive sections here and here. Remember that these values, 0 and 4, were where the expression was actually equal to 0, right here. So we do want to include those values in our solution. So notice if I start at the value 0 and then I continue to 4, and I want to include 4, and then continue further, I'm going all the way to positive infinity. So the solution for the inequality is going to be the interval from 0 to infinity, including 0. Now, what if this had been a strict inequality? What if instead of having greater than or equal to 0, what if it had read x times x minus 4 squared is strictly greater than 0? I would have exactly the same sign chart. So notice that the symbol for the inequality is not going to be what determines what the sign chart looks like. It is going to determine how we read off the answer. The answer to this one cannot include the 0 and cannot include the 4, but I still need these two intervals. So I'm going to write those answers down, those intervals down separately. The first interval starts at 0 and goes to 4, and then I want to combine that with the interval where it starts at 4 and goes to infinity. This would be the solution for the strict inequality, whereas the inequality that includes 0 I could put all together into one inequality symbol, or sorry, one interval symbol. Now, I want to go back to the chart just for a second to point out something. The first factor is linear. Notice that it's going to change signs at 0. So if I pick a number larger than 0, it's positive all the time. If I pick a number smaller than 0, it's going to be negative all the time. The second factor is a quantity that's squared. And of course, we're not including the value where it equals 0, because that's one of our cut-off points. Instead, we just notice that no matter what, if I choose a number and I square it, other than 4, it's going to be a positive result. So another way to fill in the chart is just to notice those things going on with the factors. I could fill in an entire row at a time, instead of having to do it by columns. Now, we've created this sign chart based upon an inequality that had the greater than or equal to symbol. But it turns out that the inequality symbol is not important as far as creating the sign chart. It's only reading off the answers that gives us an important part that we have to consider. For instance, what if this had been a strict inequality, as in x multiplied times x minus 4 quantity squared is greater than 0? Then I would still want to pick up the two intervals where this expression is going to be positive, but I would not want to include the endpoints. That is, I'm going to end up with two intervals. The first interval will begin at 0, and I'll put a comma and then stop at 4. And then I'll have that combined with another interval that begins at 4 and goes all the way to infinity. So the solution to the strict inequality would be the interval from 0 to 4 exclusive union the interval that begins at 4, not including 4, and goes all the way to infinity. So one more thing I want to point out before we take off on another example is if I look at what was happening with the factors, you'll see that the x value, which is linear, is going to change sign one time. It's going to change sign wherever it was equal to 0. So it's going to be positive for all values that are greater than 0, but negative for all values that are less than 0. For the second factor, x minus 4 quantity squared, that's going to be a positive value in all three intervals. Notice we're not including the value of 4. That's one of the cut-offs for the intervals. So I'm going to end up with plus signs going across in my sign chart. So you could actually fill out your sign chart, instead of filling it out in columns, you could fill it out in rows, taking advantage of what you know about those individual factors. Next we're going to look at a rational inequality. Here's our example. We have x squared plus 4x plus 16 in the numerator divided by x plus 5 in the denominator. And that whole rational expression needs to be greater than or equal to 4. Now, there's a couple of things to point out. First of all, notice that we cannot allow our x value to equal negative 5. So we always want to take note of any kind of restrictions we're going to have. So x cannot equal negative 5. Next, we have to think about the fact that if we're doing a sign chart, it always relies on these expressions being set in relation to 0. So our first step for this rational inequality will be to subtract 4 from both sides of the inequality. So we're going to have x squared plus 4x plus 16, divided by x plus 5, minus the value 4. And that's going to be greater than or equal to 0. Well, one more thing to consider is my sign chart relies on having factorization. That is, we're looking at the rules for signed numbers, where they're being multiplied, and in this case, divided. So I cannot have the subtraction of 4. I need to rewrite this expression as a single rational expression. So that means I'm going to have to get a common denominator, which is going to be my x plus 5. And what I'll do in the next step is rewrite it as x squared plus 4x plus 16 divided by x plus 5 minus-- that value of 4 I want to think of as being 4 over 1. And then we can multiply both numerator and denominator by x plus 5. I'll go ahead and put it in parentheses here. And that is now going to be greater than or equal to 0. Let's go ahead and simplify. And as we're doing that, notice my denominator will be the common denominator of x plus 5. So in my numerator, I'll have x squared plus 4x plus 16 minus 4 times the quantity x plus 5. That's being divided by now my common denominator of x plus 5. And that whole expression is greater than or equal to 0. Let's go another step. We need to distribute the 4 to give us in the numerator x squared plus 4x plus 16, now distributing the multiplication by negative 4. We have negative 4x and then minus 20. That whole thing is being divided by x plus 5. And that whole thing has to be greater than or equal to 0. Notice now that we can simplify. In the numerator, we have 4x minus 4x, which we'll end up subtracting out. We also have 16 minus 20, which will end up giving us negative 4. So the numerator condenses to x squared minus 4 divided by in the denominator x plus 5 greater than or equal to 0. I'm going to factor that numerator. We're going to have x minus 2 times x plus 2, that whole thing divided by x plus 5, greater than or equal to 0. Now, the next thing we have to do is think about in our sign chart that we're interested in where those factors are going to be changing signs. And I'm going to think about the numerator as having two factors, x minus 2, x plus 2. So that would give me values of positive 2 and negative 2 as places where those factors would change sign. In the denominator, we notice that that factor will change sign where I have x equals negative 5. So we're going to take those values, and those will be the ones that we will transfer over to a number line. Now here's an inequality for you to try on your own. The inequality is 4 divided by x minus 1 is less than or equal to 3. Pause the video and try this, and then restart it, and we'll see how you did. OK, great. You're back. Let's see what you were able to do with this. The first thing to notice is that it's not set in relation to 0. So we want to begin by subtracting 3 from both sides of the inequality. 4 divided by x minus 1 minus 3 is less than or equal 0. Now, one thing to point out is you cannot choose to multiply both sides of this inequality by the denominator to start with, because when we use multiplication in an inequality, we have to know whether or not the expression is positive or negative. So because we have a variable, we don't know whether that expression is going to be a positive value, in which case you would leave the inequality symbol as is, or whether you would be multiplying by a negative, in which case you would have to switch the inequality. That's why we have to do the subtraction of 3. We now need to get a common denominator, which will be x minus 1. So this will be 4 divided by x minus 1 minus-- we'll write it as 3 times-- so we can think of this as 3 over 1 times x minus 1 over x minus 1, less than or equal to 0. Now I have a common denominator. Let's go ahead and combine our numerators. 4 minus 3 times x minus 1 divided by x minus 1 is going to be less than or equal to 0. And now we'll go ahead and simplify the numerator. Before I continue on, I want to point out a mistake that I've seen before with students, and I want to make sure you're not making that mistake. Don't think that you can cancel the x minus 1's here. This is not a factor of the numerator. And in fact, what you'd be doing is like undoing what you had previously, only in an incorrect manner. You can't just end up with 4 minus 3. You need to recognize that you have to simplify the numerator first. Again, x minus 1 is not a factor of the numerator, and that's why it cannot be divided out. So simplifying gives us 4 minus-- we'll have 3x plus 3 divided by, in the denominator, x minus 1, and that is less than or equal to 0. Simplifying, 4 plus 3 gives us 7. 7 minus 3x is in the numerator. And x minus 1 is in the denominator. That fraction needs to be less than or equal to 0. As we did in previous examples, we want to set both the numerator equal to 0 and the denominator equal to 0 to find out where those particular factors would end up changing sign possibly. So my first 7 minus 3x set equal to 0 gives us 7 equals 3x. Dividing by 3, we get 7/3 equals x. The denominator set equal to 0 gives us x equals 1. So I need to place these two numbers on the number line to create my sign chart. Now, which one is smaller? Well, x equal to 1 is going to be the smaller value, because 1 is the same as 3/3. And 3/3 is certainly smaller than 7/3. So over to my sign chart, I'm going to place the 1 to the left of the 7/3. Set up my intervals, and put my values, or my numerator. That's 7 minus 3x and x minus 1 from the denominator. We're now going to look at that quotient once we figure out the signs of each of these. Let's choose a value larger than 7/3, so for instance, 9/3, which is going to equal the value 3. 7 minus 3 times 3 is going to be 7 minus 9, which is negative. 3 minus 1 is going to be positive. Choosing a value between 1 and 7/3. Well, let's remember that 1 is equal to 3/3. So if you just move up by thirds, you have 3/3, 4/3, 5/3, 6/3 is smaller than 7/3, and 6/3 can be reduced or rewritten as the number 2. Makes it easy to work with. So 7 minus 3 times 2 is going to be 7 minus 6, which is going to be positive. The value 2 minus 1 is still going to be positive. And finally, choose a number smaller than 1 for the interval to the left. That's easy to use the value 0. 7 minus 3 times 0 is positive. 0 minus 1 is negative. And now we can look at what happens if we have a negative number divided by a positive number. The result is negative. A positive number divided by a positive number would be positive. And a positive number divided by a negative number is negative. So we're able to fill in our sign chart using the signs of the numerator and denominator. In this example, we're interested in where that fraction is less than or equal to 0, which means we're interested in the negative intervals. This interval begins from the left at negative infinity, and continues until I get to the number 1. Now, can I include one in my solution set? The answer is no. If we look back at the outset of the problem, we notice that x minus 1 is coming from the denominator. And so therefore, we do not want to ever include that value in our solution set. And then we need this second interval, which is going to begin at 7/3. Now, 7/3 is a value that comes from the numerator. We want to allow the fraction to equal 0. And therefore, 7/3 makes the numerator 0. And that's part of our solution set. So we're going to use a bracket at the 7/3, and then continue that interval to infinity. So the solution set to this problem is negative infinity, 1, with parentheses union bracket 7/3, infinity. I hope you did well on the problem, and it's time now for you to try more on your own. Good luck. It's time for another quick quiz. What is the solution to the inequality given by 5 divided by the quantity x squared plus 1 is less than 0. Is the correct answer? A. Parentheses negative 1 comma 1 close parentheses. B. Open bracket negative 1 comma 1 closed bracket. or C. The null set. Choose A, B, or C. You're right, the answer is C, the null set. Sorry the answer is C, the null set. Let's consider the fraction where we have the value of 5 in the numerator and the denominator x squared plus 1. We want that fraction to be less than 0 which means we want the fraction to be negative. Right now it's because the numerator is a constant value of 5 we know that the numerator is a positive value. In order for the entire fraction to be negative that would require our denominator to be negative so we would have a positive number divided by a negative number. Well this means that x squared plus 1 would have to be less than 0 which means that x squared would have to be a value less than negative 1. For real number values we know that when we square a number we're going to end up either with a positive value or 0 so there's not going to be a real number value where squared is going to be less than a negative 1. Hence the solution here is that there is no solution, the null set is the correct answer.