WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take a look at some polynomial inequalities.
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When it comes to polynomial inequalities, I’m going to show you two techniques that you can use to handle these.
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One, you can take a look and figure out their graph and figure out where they are greater than or equal to 0
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or you can use a table to tracked down their sign.
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Both of these are important and since we have developed lots of graphing techniques they are both very handy.
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A polynomial inequality is simply an inequality involving a polynomial of some sort.
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Down below I have many different examples of what I'm talking about.
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Here is a nice -2x⁴ - x² - 2 > 4 that will be an example of something we are trying to solve.
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With this one, 8x³ + 2x² < 6x + 5 is another good type of inequality that we could possibly find the solution to.
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Depending on the type of polynomial present, sometimes you can call these linear inequalities, quadratic inequalities or even cubic inequalities.
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I will not get too technical usually we will just call these like polynomial inequalities.
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To solve one of these inequalities, it is best to get it in relation to 0.
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It means get everything over onto one side of your inequality symbol.
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The reason why this is so important for the entire solving process is we are looking for those ranges of values that make the inequality true.
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When it is all in relation to 0 then we just have to know where it is going to be positive or negative
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rather than trying to search for when it is greater, less than a some other number like 2 or 3.
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That is much more difficult to do.
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If you have a graph of the polynomial, this makes it especially easy because then you are essentially looking at whether it is above or below the x axis.
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Let us take a linear equality like this.
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We might know where it crosses the x axis everywhere where it is above the x axis we know it is positive.
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Everywhere below the x axis we know the value is negative.
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If I had the equation for this and maybe I wanted to know where it was greater than 0
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then I can simply take all of the positive values where it is greater than the x axis.
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Let us try a quick example to see this in action.
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I want to use the graph of x² - 3x -4 to see where it is greater than 0.
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It is already said in relation to 0 for us, we do not have to work there.
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I just need a good rough sketch of what this looks like.
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I can figure out where this thing is equal to 0 by using some of our factoring techniques.
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Our first terms better be x and then - 4 + 1 will do fine.
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I know it crosses the x axis at 4 and -1.
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Since we have also done a lot with graphing quadratic like this one, we know which direction it is facing.
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The (a) value is positive, so it is facing up.
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Let us see what else do we know about this?
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It looks like it has a y intercept of -4.
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Let us put that in there.
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A rough sketch of this might look something like that.
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You will note that I did not go through all of the work to find the vertex because I’m not interested where it reaches its maximum or minimum value.
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I'm interested in where is it above or below the x axis.
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For this particular polynomial we want to know where it is greater than 0 because that is a way of saying where does it take on positive values.
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We can see that it does this after the value of 4 and does it before the value of -1.
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In those intervals, if you are using saying those x values, then the equation will be positive.
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We will take those intervals as our solution, from negative infinity up to -1, and from 4 up to infinity.
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That grabs both of those intervals.
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We will use out little union symbol in between the show that it could be in either one of those.
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That was a quadratic example but potentially you could use these for some much larger polynomials.
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The key is creating an accurate enough graphs so we can see whether it is above or below the x axis.
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You want it in relation to 0 to see what it is doing.
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I’m just going to give a rough sketch of a polynomial, so you can see how this process might work.
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Let us call this my polynomial and I want to know where this one is less than or equal to 0.
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Everywhere it is above the x axis would be positive and everywhere below the x axis it takes on negative values.
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Look at these portions right here, here is below the x axis.
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It has another little part right there and this last term.
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We can highlight the intervals where you take on these negative values.
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Here, here and here.
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Okay, now we just have to describe those intervals using our interval notation.
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It is below the x axis from negative infinity all the way up to 1, 2, 3, 4, 5, -5 then it goes from -1 up to 3.
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I got the 3, 4, 5 from 5 up to infinity.
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We will use our little union symbol to connect all of those intervals.
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One quick note, we are including these end points because it takes on the value of 0 at these points.
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We want them included.
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Having a graph is handy.
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It gives us a nice visual way to see what is going on whether it is greater than 0 or less than 0,
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but unfortunately we do not always have a graph to work with.
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I’m thinking of some much more complicated polynomials.
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In situations like this, we are going to use a table to keep track of the sign of its factors.
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That way we will still be able to figure out whether it is above or below the x axis without ever seeing the graph.
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Here is how this process works.
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We will first get the inequality in relation to 0, so we will get everything over onto one side.
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Then we will go ahead and solve the related equation.
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Instead of looking at the inequality we will throw in an equal sign in there and then solve it and figure out where it is equal to 0.
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The reason why we are doing this is we want know where it could possibly change sign.
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Where does it hit the x axis?
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That will allow us to divide up the x axis into a lot of different intervals using those points.
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We will use our values around those points to see whether it is positive or negative in those factors.
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Once we have a lot of information about where is positive or negative
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we will determine which of those intervals actually satisfy the inequality.
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We will check the end points of those intervals to see if we need to include them or not.
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If we see something like or equals, we will usually include the values of our end points using brackets.
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If it is a nice strict inequality, then we will usually use parentheses to say that those end points are not included.
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Let us borrow a problem that we did earlier for quadratic and see how the same process works out now using a table.
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The first thing we want is we want it in relation to 0, which fortunately it is already is.
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We do not have to worry about that part.
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I’m going to look at the corresponding equation.
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Where could this thing equal 0?
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I think we factored this one earlier it is going to come in handy once more.
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I have x - 4 and x + 1.
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There are two values this could equal 0, at 4 and -1.
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Here is where our table comes into play.
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I’m going to make a number line and I’m going to mark out the location of -1 and the location of 4.
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You can see that t does split up our number line into these different intervals.
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I need to figure out what is the polynomial doing on these different intervals.
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I’m going to write down the factors of the polynomial and just experiment
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to see whether they are ending up being positive or negative on these intervals.
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That is what our test points are for.
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Let us give this a try.
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I’m going to choose something on my number line that is less than -1.
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Think of something like -2, something on that side.
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We are going to test it into these factors over here.
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If I was to plug in -2 in for x and then I subtract a 4, will that give me a positive value or negative value?
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-2 -4 will be -6 and I know on this interval that factor is a negative.
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What if I took -2 and I put in the second factor, and I added 1 to it.
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-2 + 1 I can see that that give me -1.
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Of course, the most important part is that factor would be negative on that interval.
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I’m keeping track of the individual parts of the polynomial seeing whether positive or negative.
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Now that we have a good idea what is going on that interval, let us pick something on the next interval.
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Something between -1 and 4.
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Since I have worked out good, let us test out 0 and see what it does.
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0 - 4 will that be positive or negative?
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That will give us -4.
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Let us plug in 0 into x + 1 and that will give us 1.
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I know that it is positive on that interval.
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Now something larger than 4, let us test out something over there.
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Let us test out something like 5.
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We will put that into each of our factors, 5 - 4 will give us 1 and 5 + 1 will give us 6.
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We have lots of information about what each of these individual factors are doing on these intervals.
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Of course, if we look back at the original problem that we are looking at, these factors are multiplied together.
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Let us multiply these individual pieces together and least their signs and see what is happening to the overall polynomial.
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If I were to take a negative × a negative, I would get a positive value.
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If I were to take a negative × a positive and I will get a negative value.
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If I multiplied two positives together, I will get a positive value.
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This is the sign of the overall polynomial when I put those pieces together.
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What are my interests in, positive or negative values?
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I'm looking for when the polynomial is greater than 0, so I want those positive values.
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We will write down the value of the intervals where it was positive.
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The things greater than 4 and things less than -1.
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Note how I'm not including the end points here because we are dealing with a strict inequality.
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This would represent the solution to our polynomial inequality.
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Of course we might be doing with some more complicated polynomials and that is why we are using these techniques of a table.
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I have 4x³ - 7x is less than or equal to 15x.
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I have picked up a lot of the techniques for graphing this in a nice easy way other than just making a giant table of values.
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Let us use our table here, we would be able to figure out what the interval or range of solutions is for this one.
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I'm going to get everything over onto one side first.
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It looks like I'm looking where this is less than or equal to 0 or below that x axis.
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We want to solve where this thing is equal to 0.
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We will have to borrow some of our factoring techniques to help out.
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Everything has an x in common.
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Now we got all of those taken out.
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We can go ahead and factor the remaining quadratic.
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What values could I use in here?
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I have a 4x, 1x, 5, -3.
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First terms will give us 4x², outside would be -12, inside would be 5 and our last terms -15.
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This checks out.
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x = 0, 4x + 5= 0 and x – 3 = 0.
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We would end up solving each of these separately.
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This middle one, let us go ahead and move the 5 over and then divide by 4.
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For the third one we will simply add 3 to both sides.
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I have these 3 points where our polynomial could end up changing sign because that is where it is equal to 0.
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Let us go ahead and draw our number line and put these values on there.
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That way we can see how it divides up our number line.
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The smallest thing we have here is -5/4.
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The next largest would be 0.
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The largest value we have is 3.
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Always put these from smallest to largest.
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Let us also keep track of the pieces and our polynomials.
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We have x, 4x + 5 and we have x – 3.
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We are going to use our test value into these individual parts and put them together in the end.
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Let us start off with our first test value.
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We need something that is less than -5/4.
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I know -5/4 is close to -1 I’m going to choose something a little bit farther.
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It looks like I’m going to choose -2 and put it into all of our factors.
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If I put in -2 in 4x I will get -2.
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If I put in -2 in for 4 then 4x + 5 that will give me -x + 5 which would still be negative.
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Last that I could take that -2 and put it down into x -3, it will give me -5.
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Negative again.
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Let us choose something between -5/4 and 0.
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Here is where that -1 will come into play.
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Putting a negative 1 in for x will give us a negative value.
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We need in 4 for 4x + 5 will give us 1.
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Putting in for the last one, a -1 -3 = -4 so minus down there.
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That takes care of that.
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Let us see what happens when we put in 2 for all of these spots.
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2 and 4x that will be positive.
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Putting in 2 for 4x + 5 that will be 8 + 5 =13.
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If I put in 2 for the last one, that will be 2 -3 = -1.
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Negative.
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One more last thing, things that are greater than 3.
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I’m going to test that out with 4.
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Let us see if we can put that in all the spots.
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I’m putting in for x and I will get 4 × 4 + 5 = 21 and 4 - 3 =1.
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There would be that one.
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That represents all of the individual components in the polynomial and what their sign will be on those intervals.
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We want to imagine all of these being put together since all of these factors are being multiplied.
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Let us do this one column at a time.
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Negative × negative × another negative, that value would be negative.
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We have a negative × a positive and that will be negative, times another negative and that would be positive.
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Now we have a positive × positive × negative, that is negative.
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Our last column are all positives being multiplied together so positive.
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I have a good sense of what the overall polynomial is doing for these different intervals.
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I can see when it is negative, positive, negative and positive.
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This problem right here, we want to know whether it is less than or equal to 0.
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We are looking for the negative intervals.
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You have this interval down here and we have this interval right here.
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We can go back and highlight those on a number line.
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I'm looking at the values all the way up to -5/4 and I’m looking at the values between 0 and 3.
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Let us write down those intervals.
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negative infinity up to -5/4 and from 0 to 3.
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Note that I'm definitely including my end points here because this says or equal to 0.
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I'm also interested in where my polynomial is equal to 0.
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This would be our intervals solution.
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No matter what method you use, it is often very important that you know where your polynomial is equal to 0.
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Use many of our factoring techniques to help you out for that.
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Then you can go ahead and look at the graph or even one of these tables to organize your information.
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That will make your life much easier.
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Thank you for watching www.educator.com.