WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In a previous lesson, we talked about solving quadratic equations by graphing.
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And we are going to go on to discuss another technique, an algebraic technique, which is solving quadratic equations by factoring.
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So, first I am starting out with just a brief review of the different factoring techniques.
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And each of these is covered in-depth in the Algebra I series.
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So, if you are unsure on any of these, make sure you review that before you go on,
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because we are just using these techniques as a tool, now, to actually solve quadratic equations.
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OK, first, remember the greatest common factor: the **greatest common factor** of two or more numbers is the product of their common prime factors.
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So, the GCF of two or more numbers is the product of their common prime factors.
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And, in factoring, often, when you are factoring an equation, if there is a GCF, you want to factor that out first,
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because then you are working with something less complicated.
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For example, if you had 4x² + 16x + 36, you will recognize that there is a GCF of 4.
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So, I am going to factor that out and get 4 times x squared, plus 4x, plus 9.
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Factoring out the 4 gives me x² + 4x + 9; and this is much easier to work with.
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So, remember to factor out the GCF first, if there is one.
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Recall the difference of two squares: you will recognize this, because it is going to be in the form a² - b².
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So, an example would be x² - 4; and that is equal to (x + 2) (x - 2).
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So, this is the difference of two squares; here is the plus and the minus.
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And so, when you are factoring, if you recognize this, you can quickly factor it as the difference of two squares,
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knowing that it fits into this formula, a² - b².
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Perfect square trinomials: **perfect square trinomials** would be the result of squaring a binomial.
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For example, if you have x² + 6x + 9, this is actually equal to (x + 3)².
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So, if you take a binomial and square it, you will get a perfect square trinomial.
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Take a binomial; multiply it times itself; that is what a perfect square trinomial is.
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And so, recognizing that makes for easy factoring.
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General trinomials are a little more complex to factor; again, you can review all of this in Algebra I--techniques for factoring.
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And an example of a general trinomial would be x² + x - 12;
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it doesn't fit into any of these special cases, like the difference of two squares, or perfect square trinomials.
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So, I have to do a little more work in factoring that out.
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Recall from earlier lessons that you need to look at the signs.
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I have a positive sign here; I have a negative here.
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And if I have a negative, that must be the result of multiplying a positive and a negative.
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I then look at what the factors of 12 are; and the factors of 12 are 1 and 12, 2 and 6, and 3 and 4.
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And one will be positive, and one will be negative; and I need to have factors that sum to this middle term, which is actually 1.
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And I can try different combinations; and I know that -1 and 12 is not going to work; -12 and 1 certainly will not work.
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And I go on down, and from that, you can see that, if I want them to sum to 1, I am going to need factors that are close to each other.
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And the closest two I have here are 3 and 4.
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So, if I make 4 positive and 3 negative, then when I add the outer term and the inner terms, I will get my middle term x.
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And you can always check this using the FOIL method: First terms (that is x²),
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Outer terms (-3x), Inner terms (4x), and then the Last is -12.
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Simplifying this, I get my original back.
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General trinomials take a little more work, and you can always check those by multiplying it back out to make sure you did it correctly.
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So, we are making sure you have all these and know how to use them well.
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And then, we are going to be using them today to actually solve some quadratic equations.
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Now, once you have factored, you need to use the zero product rule to actually find the solutions.
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And what the zero product rule says is that, for any number a and b, if ab is zero, then either a is zero or b is zero--
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because if a equals 0, then you would get 0 times b; that would work as a solution;
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if b is 0, then 0 times a would give you 0, and that is also a solution.
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For example, if I was given x² - 16 = 0 and asked to solve that, I would first recognize that it is
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in the form a² - b², and that it is therefore the difference of two squares.
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That allows me to factor it pretty quickly into (x + 4) (x - 4).
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So, I am factoring this, and it still equals 0.
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Now, use the zero product rule: the zero product rule tells me that, if this is 0 or this is 0, then this entire thing will equal 0, which is what I want.
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So, if x + 4 equals 0, this will be solved; if x - 4 equals 0, this will be solved.
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So, I am going to set this factor equal to 0 and solve for x; I am going to set this factor equal to 0 and solve for x.
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And that is going to give me...let's see...x = 4.
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And if you wanted to check that, you could go back up here and say, "OK, let x equal -4."
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So, that is -4² - 16 = 0; that is 16 - 16 = 0, and that checks out; that is a valid solution.
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I could do the same thing for 4; x = -4, or I could say x = 4; and that is going to give me 4² - 16 = 0.
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16 - 16 does equal 0, so there are two solutions here.
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And I was able to find those using factoring and the zero product rule.
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So, trying this out: first I am just asked to factor; and recall that the first thing you want to do is factor out any greatest common factor,
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because that is going to make whatever is left much easier to work with.
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And I see that I have a greatest common factor of 4.
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This factors into 4 times x² - x - 6.
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Now, all I have here is a general trinomial, so I want to think about what I have.
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And I want to make sure that I bring my 4 along with me, because that is part of the solution.
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I have a negative sign here; and the only way you are going to end up with that is if one of these is positive and one is negative.
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Now, I am going to think about what my factors of 6 are; I have 1 and 6, and 2 and 3.
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I need factors that sum to a middle term of -1; and that is not a very large number, so I am going to look for factors that are close together.
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I am going to try these first: now one is negative, and the other is positive.
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Since this is negative, I am going to look for making the larger number negative; so let me try if I have -3 plus 2.
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That equals -1, and that gives me the coefficient of that middle term; so this is what I have.
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And I can always check that by using FOIL to go back and multiply these out;
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and then I would have to multiply the 4 back into it.
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But this is factored; so I first factored out the GCF, and then I saw that I can take it farther, because this is not factored out all the way.
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It is a general trinomial that factors into (x + 2) (x - 3), times that greatest common factor, 4.
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OK, now we are asked to actually solve; and this is 3x² = 27.
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Before you can solve a quadratic equation, you need to put it in standard form.
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Recall that standard form is ax² + bx + c = 0.
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So, I have 3x² = 27; to put this in standard form, I am going to subtract 27 from both sides.
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And I have an ax² term; b must be 0, because there is no x term; and then I have a c of -27.
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So, I am going to solve by factoring; I have this in standard form--now I am going to factor out the GCF.
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And the GCF is 3; I am going to pull that out.
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Now, I am looking at this, and I see that what I have is something in the form a² - b², which is the difference of two squares.
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Always make sure you bring this GCF down--don't leave it behind.
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(x - 3) (x + 3) = 0; so now, I am going to use the zero product property (or zero product rule) to find my solutions.
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And the zero product rule says that if a times b equals 0, then a equals 0 or b equals 0, or they both have to be 0.
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So, first I am going to have 3 times (x - 3) equals 0; and if I divide 0/3, I am just going to get x - 3 = 0, or x = 3.
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So, that is one solution; the other solution is going to be x + 3 = 0.
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Using the zero product rule, that tells me that x equals -3; so my two solutions are that x equals 3 and x equals -3.
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These are my two solutions for this quadratic equation.
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And I found that just by factoring: x = 3 and x = -3; and I made my factoring a lot easier by first pulling out the GCF.
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Again, solve by factoring: and we have another situation where it is not in standard form.
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So, I am going to put it in standard form, which is ax² + bx + c = 0.
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All right, that is going to give me 4x² - 24x + 36 = 0, just subtracting 24x from both sides.
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This is another situation where I have a greatest common factor; so I have a GCF equal to 4--factor that out.
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Pull that out: that is 4, times x² - 6x + 9, equals 0.
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Again, I have something much easier to work with since I have pulled that out.
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So, figuring out how to work with this, I am going to go ahead and factor this out, because it is not all the way factored.
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And here, I have (x - 3) times (x -3); and it is actually a perfect square trinomial; this is really just (x - 3)².
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OK, so this is a perfect square trinomial, because you have x², and then (just check this using FOIL)
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I would have x² - 3x - 3x + 9; so I know I did that correctly.
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Now, I am going to use the zero product rule, which tells me that, if one of these is equal to 0, then the entire thing is equal to 0.
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So, I can use that to find the solution.
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Now, you can just go ahead and divide both sides by 4, in which case this will move over to here, and that just stays 0.
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Really, I just need to work with this and this: x - 3 = 0, and this is the same thing: x - 3 = 0.
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So, actually, when I figure this out, I just get the same thing for both; and it is x = 3.
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So, I only have one solution, or one real root, in this case; so, x equals 3.
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And again, put it in standard form; factor out the greatest common factor; complete your factoring;
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and then use the zero product rule to determine that there is one real solution, and that is that x equals 3.
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OK, again, solve by factoring; this is already in standard form; however, I have to get rid of this greatest common factor, which is 2.
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So, I am pulling that out to get 2x² + 5x - 12.
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When the leading coefficient is not 1, the factoring is a bit more difficult.
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So, let's move over here and work on this.
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When the leading coefficient is 2, I am going to have something in this form.
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And I have a negative here, so I also know that one of these is going to be a positive, and one is going to be a negative.
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But I don't actually know which one yet; but I know I am going to have +/-, or I am going to have this.
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OK, let's look at some factors of 12: factors of 12 would be 1 and 12, 2 and 6, and 3 and 4.
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Now, 5 is not that large of a number, especially when we think about the fact that we are going to have to be also multiplying by 2.
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So, if I go and use something like 6, and it ends up getting multiplied by 2, it is going to be very large; the difference is going to be great.
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I want factors that are smaller, since the difference between those, even with this 2x thrown in, is only going to be 5x.
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So, I am going to start with these, because they don't have a large difference between them.
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So, I am going to start out just trying (2x + 3) (x - 4) and seeing what I get.
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I am not worried about the first term; I am worried about the outer terms added to the inner terms.
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And see if I get the correct middle term.
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I am looking for the middle term equal to 5x.
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This is going to give me 2x times -4; that is -8x, plus 3x; that is 5x.
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I have the right idea, but I actually have the wrong signs here.
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So, I am going to try reversing the signs, because (this is -5x) I want this to be 5x, because here I have -8x + 3x is -5x.
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So, the same idea: let's try different signs, though.
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This time, I am making this negative and this positive; this is going to give me 2x(4); that is going to give me +8x; -3x is 5x.
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So, this is correct; I got the correct middle term, so this is the correct factoring.
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And this can be a lot of work to factor these; so it is important to go logically--
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for example, seeing that I don't have a very large term here, especially when I am dealing with the 2x also
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(it is going to amplify things), to look for factors that are not very far apart.
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OK, so now, I am back here, and I am solving by factoring.
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This is going to give me (2x - 3) (x + 4) = 0.
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Dividing both sides by 2, this 2 is just going to drop out.
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So, when I use the zero product property, I am going to get 2x - 3 = 0, and I am also going to get x + 4 = 0.
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So, I just need to go ahead and solve those to get 2x = 3, or x = 3/2.
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Here, I just have x + 4 = 0, and that is simple: it is x = -4.
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I have two solutions: x = 3/2 and x = -4.
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I solved this by pulling out the greatest common factor, factoring that out, then factoring this trinomial into this,
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and using the zero product rule to give me 3/2 for a solution from here, and x = -4 for a solution from here.
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Thanks for visiting Educator.com, and I will see you next lesson!