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Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | The roots are also evident when we look at this graph here. We have a root at x equals negative four, a root at x equals negative 1 1 1 1 1, or negative 3 1 1 1 1 1, and a root at x is equal to one. So really what we have to do is say, which of these factors are consistent with the roots that we see? So let's go root by root. So here on the left, we have a root at x equals negative four. In order for this polynomial to be zero when x is equal to negative four, that means that x plus four must be a factor, or some multiple, or some constant times x plus four must be a factor of our polynomial. Now we can see in the choices that we have a bunch of x plus fours, but they have different exponents on them. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | So let's go root by root. So here on the left, we have a root at x equals negative four. In order for this polynomial to be zero when x is equal to negative four, that means that x plus four must be a factor, or some multiple, or some constant times x plus four must be a factor of our polynomial. Now we can see in the choices that we have a bunch of x plus fours, but they have different exponents on them. The first one has a two as an exponent, it's being squared, while the others have a one as the exponent. Now what we've talked about in other videos when we talked about multiplicity, we said, hey, if we see a sign change around a root, like we're seeing right over here around x equals negative four, that means that we are going to see an odd exponent on the corresponding factor. But if we didn't see a sign change, as we see in this other root over here, that means we would see an even exponent. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | Now we can see in the choices that we have a bunch of x plus fours, but they have different exponents on them. The first one has a two as an exponent, it's being squared, while the others have a one as the exponent. Now what we've talked about in other videos when we talked about multiplicity, we said, hey, if we see a sign change around a root, like we're seeing right over here around x equals negative four, that means that we are going to see an odd exponent on the corresponding factor. But if we didn't see a sign change, as we see in this other root over here, that means we would see an even exponent. Now we clearly see the sign change, so we would expect an odd exponent, and of course, one is an odd number and two isn't. So if you just have a straight up x plus four, you would have a sign change around x equals negative four. So I can rule out this first choice. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | But if we didn't see a sign change, as we see in this other root over here, that means we would see an even exponent. Now we clearly see the sign change, so we would expect an odd exponent, and of course, one is an odd number and two isn't. So if you just have a straight up x plus four, you would have a sign change around x equals negative four. So I can rule out this first choice. These other three choices are still looking good based on just the x plus four factor. Now let's move on to the next factor right over here, so, or the next root. The next root is at x is equal to negative 3 1 2, and so one way to think about it is you could have a factor that looks like x plus 3 1 2, or this times some constant. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | So I can rule out this first choice. These other three choices are still looking good based on just the x plus four factor. Now let's move on to the next factor right over here, so, or the next root. The next root is at x is equal to negative 3 1 2, and so one way to think about it is you could have a factor that looks like x plus 3 1 2, or this times some constant. Now when we look at the choices or the remaining choices, we don't see x plus 3 1 2, but we do see something that involves a two and a three, and so one way to think about it is, hey, if I just multiply this times the constant two, so that would get us two x plus three, well, I do see that right over here, and then the next question is what should be the exponent? Well, once again, we have a sign change around x equals negative 3 1 2, so we would expect an odd exponent there, and you can see out of the choices, only two of them have an exponent of one, which is an odd number, while the other one has an even exponent there, so we can rule that one out as well, and then we go to this last root. I will do this in an orange color. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | The next root is at x is equal to negative 3 1 2, and so one way to think about it is you could have a factor that looks like x plus 3 1 2, or this times some constant. Now when we look at the choices or the remaining choices, we don't see x plus 3 1 2, but we do see something that involves a two and a three, and so one way to think about it is, hey, if I just multiply this times the constant two, so that would get us two x plus three, well, I do see that right over here, and then the next question is what should be the exponent? Well, once again, we have a sign change around x equals negative 3 1 2, so we would expect an odd exponent there, and you can see out of the choices, only two of them have an exponent of one, which is an odd number, while the other one has an even exponent there, so we can rule that one out as well, and then we go to this last root. I will do this in an orange color. We have a root at x equals one, so we would expect x minus one, or this multiplied times some constant, to be one of the factors, and what's interesting here is we don't see a sign change around x equals one, so we would expect an even exponent, and so out of the remaining choices, we see an x minus one in both of them, but only choice C has the even exponent that we would expect, so choice C is looking good. If we were to go with choice D, where this is to the first power, we would expect a sign change around x is equal to one, so this would be a situation where the curve would keep going down, something like that, so we like choice C. Let's do another example. So once again, we are asked what could be the equation of P, and we're given a graph, so again, pause this video and try to work through that. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | I will do this in an orange color. We have a root at x equals one, so we would expect x minus one, or this multiplied times some constant, to be one of the factors, and what's interesting here is we don't see a sign change around x equals one, so we would expect an even exponent, and so out of the remaining choices, we see an x minus one in both of them, but only choice C has the even exponent that we would expect, so choice C is looking good. If we were to go with choice D, where this is to the first power, we would expect a sign change around x is equal to one, so this would be a situation where the curve would keep going down, something like that, so we like choice C. Let's do another example. So once again, we are asked what could be the equation of P, and we're given a graph, so again, pause this video and try to work through that. All right, we're gonna do the same idea. Let's go to this first root right over here. We have a root at x equals negative three, so we would expect some multiple of x plus three to be one of the factors. |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | So once again, we are asked what could be the equation of P, and we're given a graph, so again, pause this video and try to work through that. All right, we're gonna do the same idea. Let's go to this first root right over here. We have a root at x equals negative three, so we would expect some multiple of x plus three to be one of the factors. We also have a sign change around x equals negative three, so we would expect an odd multiplicity, and we would expect an odd exponent on the x plus three factor. When we look at all the choices, C and D have an even exponent, so if we had x plus three to the fourth, then you wouldn't have a sign change here. You would just touch the x-axis and then go back to where it was coming from, so we can rule out these choices, and now let's look at the second root right over here at x is equal to two, so we would expect x minus two to be one of the factors or a multiple of this, and because we don't have a sign change around x equals two, the graph just touches the x-axis and then goes back to where it was coming from. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | 18x to the fourth minus 3x squared plus 6x minus 4. All of that over 6x. So there's a couple of ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decompose this numerator up here. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decompose this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a over d plus b over d plus c over d. Or maybe not so clearly, but hopefully that helps clarify. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | One is I just kind of decompose this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a over d plus b over d plus c over d. Or maybe not so clearly, but hopefully that helps clarify. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something. The other way to think about it is that we're multiplying this entire expression, so this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something. The other way to think about it is that we're multiplying this entire expression, so this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you. They're all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | Whatever seems to make sense for you. They're all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here we can just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here we can just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the... Well, they don't tell us, but if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the four minus one power, or x to the third power. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | 18 divided by 6 is 3. And then you have x to the fourth divided by x to the... Well, they don't tell us, but if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the four minus one power, or x to the third power. Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | That's going to be x to the four minus one power, or x to the third power. Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1 half. And then you have x squared divided by x. We already know that x is the same thing as x to the first. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | So I'm going to do this part next. Negative 3 divided by 6 is negative 1 half. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just... Well, I'll write it. I could write a 1 here. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just... Well, I'll write it. I could write a 1 here. Let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. It gives you two ways. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | I could write a 1 here. Let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. It gives you two ways. Anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, which is x to the 0, which is also equal to 1. Either way. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | It gives you two ways. Anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, which is x to the 0, which is also equal to 1. Either way. You knew how to do this before you even learned that exponent property, because x divided by x is 1. And then assuming x does not equal 0. And then finally, and we kind of have to assume x doesn't equal 0 in this whole thing, otherwise we would be dividing by 0. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | Either way. You knew how to do this before you even learned that exponent property, because x divided by x is 1. And then assuming x does not equal 0. And then finally, and we kind of have to assume x doesn't equal 0 in this whole thing, otherwise we would be dividing by 0. And then finally we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is both 4... |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | And then finally, and we kind of have to assume x doesn't equal 0 in this whole thing, otherwise we would be dividing by 0. And then finally we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is both 4... Negative 4... Let me do it. Negative 4 over 6 is the same thing as negative 2 thirds. Just simplify that fraction. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | So the simplest way is both 4... Negative 4... Let me do it. Negative 4 over 6 is the same thing as negative 2 thirds. Just simplify that fraction. And we're multiplying that times 1 over x. And so we could multiply that times 1 over x. So we could do this 4 times 1 over x. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | Just simplify that fraction. And we're multiplying that times 1 over x. And so we could multiply that times 1 over x. So we could do this 4 times 1 over x. Another way to think about it is, you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the 1st power. And then when you tried to simplify it using your exponent properties, you would have... Well, that would be x to the 0 minus 1 power, which is x to the negative 1 power. So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | So we could do this 4 times 1 over x. Another way to think about it is, you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the 1st power. And then when you tried to simplify it using your exponent properties, you would have... Well, that would be x to the 0 minus 1 power, which is x to the negative 1 power. So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. It's going to be 3x to the 3rd minus 1 half x plus 1. Because this thing right here is just 1. |
Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3 | So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. It's going to be 3x to the 3rd minus 1 half x plus 1. Because this thing right here is just 1. So plus 1. And then minus 2 times 1 in the numerator over 3 times x in the denominator. And we are done. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | So we have negative 30x to the fifth is equal to negative 10x to the third times f. And I encourage you to pause this video and see if you can figure out what f is going to be. Well, the way that we can tackle it, we could just isolate the f on the right-hand side here if we divide by negative 10x to the third. So we might say, well, we want to divide this side by negative 10x to the third. But if we want the equality to be true, if we want the left side to stay being the same as the right side, we have to do, whatever we do to the right side, we have to do to the left side as well. So we have to divide the left side by negative 10x to the third. And then what does that leave us with? Well, on the right-hand side, up top, we're multiplying by negative 10x to the third, and then we're dividing by negative 10x to the third. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | But if we want the equality to be true, if we want the left side to stay being the same as the right side, we have to do, whatever we do to the right side, we have to do to the left side as well. So we have to divide the left side by negative 10x to the third. And then what does that leave us with? Well, on the right-hand side, up top, we're multiplying by negative 10x to the third, and then we're dividing by negative 10x to the third. Well, multiplying by something and then dividing by that same thing is the same thing as just multiplying by one, or you could, one way to think about it, they just cancel out. So we are just going to be left with an f. We're going to be left with an f on the right-hand side. There's the whole point, we wanted to solve for f. And on the left-hand side, we see, we can first look at the coefficients. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | Well, on the right-hand side, up top, we're multiplying by negative 10x to the third, and then we're dividing by negative 10x to the third. Well, multiplying by something and then dividing by that same thing is the same thing as just multiplying by one, or you could, one way to think about it, they just cancel out. So we are just going to be left with an f. We're going to be left with an f on the right-hand side. There's the whole point, we wanted to solve for f. And on the left-hand side, we see, we can first look at the coefficients. We could say negative 30 divided by, negative 30 divided by negative 10 is positive three. So that's going to be three. And then x to the, x to the fifth power divided by x to the third power. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | There's the whole point, we wanted to solve for f. And on the left-hand side, we see, we can first look at the coefficients. We could say negative 30 divided by, negative 30 divided by negative 10 is positive three. So that's going to be three. And then x to the, x to the fifth power divided by x to the third power. Well, that's going to be x squared. X squared. You could either think of it in terms of our exponent properties. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | And then x to the, x to the fifth power divided by x to the third power. Well, that's going to be x squared. X squared. You could either think of it in terms of our exponent properties. We would subtract these two exponents, x to the five minus three, which is x squared, or you could say, hey, up on top, that's x times x times x times x times x. Did I say that right? Five x's. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | You could either think of it in terms of our exponent properties. We would subtract these two exponents, x to the five minus three, which is x squared, or you could say, hey, up on top, that's x times x times x times x times x. Did I say that right? Five x's. You could view it as x, let me do it in that, on top, you have x to the fifth, which is this. I always like to remind myself why the exponent properties even work. And then on the denominator, on the denominator, you have x times x times x, and these three x's are going to cancel, and you're just going to be left with x times x, which is just x squared. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | Five x's. You could view it as x, let me do it in that, on top, you have x to the fifth, which is this. I always like to remind myself why the exponent properties even work. And then on the denominator, on the denominator, you have x times x times x, and these three x's are going to cancel, and you're just going to be left with x times x, which is just x squared. So you get f is equal to three x squared. So you could write, we could write that negative 30 x to the fifth is equal to, is equal to negative 10 x to the third times f. And now we know that f is three x squared. Three x squared. |
Worked example finding the missing monomial factor High School Math Khan Academy.mp3 | And then on the denominator, on the denominator, you have x times x times x, and these three x's are going to cancel, and you're just going to be left with x times x, which is just x squared. So you get f is equal to three x squared. So you could write, we could write that negative 30 x to the fifth is equal to, is equal to negative 10 x to the third times f. And now we know that f is three x squared. Three x squared. And so another way to describe what's going on in this equation, we could say that negative 30 x to the fifth is divisible by either one of these factors, that negative 30 x to the fifth is divisible by negative 10 x to the third, or we could say negative 30 x to the fifth is divisible by three x squared, or we could say that three x squared is a factor of negative 30 x to the fifth. And the way that we can make these claims about factor and divisibility is we're dealing with, we're dealing with non-fractional coefficients right over here, and we're also dealing with non-fractional exponents right over here. So that's why we're saying, hey, these are factors, this yellow thing and this magenta thing, factors of this blue thing, or this blue thing is divisible by either one of these. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | So let's first think about what an even function is. One way to think about an even function is that if you were to flip it over the y-axis, that the function looks the same. So here's a classic example of an even function. It would be this right over here, your classic parabola where your vertex is on the y-axis. This is an even function. So this one is maybe the graph of f of x is equal to x squared. And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | It would be this right over here, your classic parabola where your vertex is on the y-axis. This is an even function. So this one is maybe the graph of f of x is equal to x squared. And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. Now a way that we can talk about that mathematically, and we have talked about this when we introduced the idea of reflection, to say that a function is equal to its reflection over the y-axis, that's just saying that f of x is equal to f of negative x. Because if you were to replace your x's with a negative x, that flips your function over the y-axis. Now what about odd functions? |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. Now a way that we can talk about that mathematically, and we have talked about this when we introduced the idea of reflection, to say that a function is equal to its reflection over the y-axis, that's just saying that f of x is equal to f of negative x. Because if you were to replace your x's with a negative x, that flips your function over the y-axis. Now what about odd functions? So odd functions, you get the same function if you flip over the y and the x-axes. So let me draw a classic example of an odd function. Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | Now what about odd functions? So odd functions, you get the same function if you flip over the y and the x-axes. So let me draw a classic example of an odd function. Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. So notice, if you were to flip first over the y-axis, you would get something that looks like this. So I'll do it as a dotted line. If you were to flip just over the y-axis, it would look like this. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. So notice, if you were to flip first over the y-axis, you would get something that looks like this. So I'll do it as a dotted line. If you were to flip just over the y-axis, it would look like this. And then if you were to flip that over the x-axis, well, then you're going to get the same function again. Now how would we write this down mathematically? Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | If you were to flip just over the y-axis, it would look like this. And then if you were to flip that over the x-axis, well, then you're going to get the same function again. Now how would we write this down mathematically? Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. So this is doing two flips. So some of you might be noticing a pattern or think you might be on the verge of seeing a pattern that connects the words even and odd with the notions that we know from earlier in our mathematical lives. I've just shown you an even function where the exponent is an even number. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. So this is doing two flips. So some of you might be noticing a pattern or think you might be on the verge of seeing a pattern that connects the words even and odd with the notions that we know from earlier in our mathematical lives. I've just shown you an even function where the exponent is an even number. And I've just showed you an odd function where the exponent is an odd number. Now I encourage you to try out many, many more polynomials and try out the exponents, but it turns out that if you just have f of x is equal to, if you just have f of x is equal to x to the n, then this is going to be an even function if n is even, and it's going to be an odd function if n is odd. So that's one connection. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | I've just shown you an even function where the exponent is an even number. And I've just showed you an odd function where the exponent is an odd number. Now I encourage you to try out many, many more polynomials and try out the exponents, but it turns out that if you just have f of x is equal to, if you just have f of x is equal to x to the n, then this is going to be an even function if n is even, and it's going to be an odd function if n is odd. So that's one connection. Now some of you are thinking, wait, but there seem to be a lot of functions that are neither even nor odd, and that is indeed the case. For example, if you just had the graph x squared plus two, this right over here is still going to be even because if you flip it over, you have the symmetry around the y-axis, you're going to get back to itself. But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | So that's one connection. Now some of you are thinking, wait, but there seem to be a lot of functions that are neither even nor odd, and that is indeed the case. For example, if you just had the graph x squared plus two, this right over here is still going to be even because if you flip it over, you have the symmetry around the y-axis, you're going to get back to itself. But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. So it's not just the exponent, it also matters on the structure of the expression itself. If you have something very simple like just x to the n, well then, that could be or that would be even or odd depending on what your n is. Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. So it's not just the exponent, it also matters on the structure of the expression itself. If you have something very simple like just x to the n, well then, that could be or that would be even or odd depending on what your n is. Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. You're going to get something like this. So you're no longer back to your original function. Now an interesting thing to think about, can you imagine a function that is both even and odd? |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. You're going to get something like this. So you're no longer back to your original function. Now an interesting thing to think about, can you imagine a function that is both even and odd? So I encourage you to pause that video or pause the video and try to think about it. Is there a function where f of x is equal to f of negative x and f of x is equal to the negative of f of negative x? Well, I'll give you a hint or actually I'll just give you the answer. |
Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3 | Now an interesting thing to think about, can you imagine a function that is both even and odd? So I encourage you to pause that video or pause the video and try to think about it. Is there a function where f of x is equal to f of negative x and f of x is equal to the negative of f of negative x? Well, I'll give you a hint or actually I'll just give you the answer. Imagine if f of x is just equal to the constant zero. Notice, this thing is just a horizontal line, just like that at y is equal to zero, and if you flip it over the y-axis, you get back to where it was before, then if you flip it over the x-axis again, then you're still back to where you were before. So this over here is both even and odd, very interesting case. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | And I always do this before I have to convert between the two. If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. |
Radian and degree conversion practice Trigonometry Khan Academy (2).mp3 | 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians. We know that there are 180 degrees for every pi radians, so we're going to get the radians cancel out, the pi's cancel out, and so you have negative 180 over 2. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | Carlos has taken an initial dose of a prescription medication. The relationship between the elapsed time, t, in hours since he took the first dose, and the amount of medication, m of t, in milligrams in his bloodstream is modeled by the following function. All right. In how many hours will Carlos have one milligram of medication remaining in his bloodstream? So m of what t is equal to, so we need to essentially solve for m of t is equal to one milligram, because m of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So m of t is, they give us a definition. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | In how many hours will Carlos have one milligram of medication remaining in his bloodstream? So m of what t is equal to, so we need to essentially solve for m of t is equal to one milligram, because m of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So m of t is, they give us a definition. Its model is an exponential function. 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | So m of t is, they give us a definition. Its model is an exponential function. 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. And so how do we solve this? Well, one way to think about it, one way to think about it, what happens if we took the natural log of both sides? And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. And so how do we solve this? Well, one way to think about it, one way to think about it, what happens if we took the natural log of both sides? And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. This is zero, that is 0.05. So I'm gonna take the natural log of both sides. So ln, ln. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. This is zero, that is 0.05. So I'm gonna take the natural log of both sides. So ln, ln. So the natural log, this says, what power do I have to raise e to to get to e to the negative 0.8 t? Well, I've got to raise e to the, this simplifies to negative 0.8 t. Once again, natural log, this thing, let me clarify, ln of e to the negative 0.8 t, this is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to to get to e to the negative 0.8 t? We'll have to raise it to the negative 0.8 t power. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | So ln, ln. So the natural log, this says, what power do I have to raise e to to get to e to the negative 0.8 t? Well, I've got to raise e to the, this simplifies to negative 0.8 t. Once again, natural log, this thing, let me clarify, ln of e to the negative 0.8 t, this is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to to get to e to the negative 0.8 t? We'll have to raise it to the negative 0.8 t power. So that's why the left-hand side's simplified to this, and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, natural log of 0.05, all of that, and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8, and so t is going to be equal to all of this business. On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | We'll have to raise it to the negative 0.8 t power. So that's why the left-hand side's simplified to this, and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, natural log of 0.05, all of that, and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8, and so t is going to be equal to all of this business. On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. So let me get a calculator out, clear it out, and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log, we get that value, and we want to divide it by negative 0.8. So divide it by, divide it by 0.8 negative, so we're going to divide by 0.8 negative, is equal to, let's see, they want us to round to the nearest hundredth, so 3.74. |
Exponential model word problem medication dissolve High School Math Khan Academy.mp3 | On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. So let me get a calculator out, clear it out, and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log, we get that value, and we want to divide it by negative 0.8. So divide it by, divide it by 0.8 negative, so we're going to divide by 0.8 negative, is equal to, let's see, they want us to round to the nearest hundredth, so 3.74. So it'll take 3.74, 7.4 hours for his dosage to go down to one milligram, where it actually started at 20 milligrams. When t equals zero, it's 20. After 3.74 hours, he's down in his bloodstream to one milligram, I guess his body has metabolized the rest of it in some way. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | This green area is 12x to the fourth, this purple area is 6x to the third, this blue area is 15x squared. You add them all together, you get this entire rectangle, which would be the combined areas, 12x to the fourth plus 6x to the third plus 15x squared. The length of the rectangle in meters, so this is the length right over here that we're talking about, we're talking about this distance. The length of the rectangle in meters is equal to the greatest common monomial factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and width of the rectangle? I encourage you to pause the video and try to work through it on your own. Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | The length of the rectangle in meters is equal to the greatest common monomial factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and width of the rectangle? I encourage you to pause the video and try to work through it on your own. Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. If the length is the greatest common monomial factor of these terms, of 12x to the fourth, 6x to the third, and 15x squared, well then we can factor that out, and then what we have left over is going to be the width. So let's figure out what is the greatest common monomial factor of these three terms. The first thing we can look at is let's look at the coefficients. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. If the length is the greatest common monomial factor of these terms, of 12x to the fourth, 6x to the third, and 15x squared, well then we can factor that out, and then what we have left over is going to be the width. So let's figure out what is the greatest common monomial factor of these three terms. The first thing we can look at is let's look at the coefficients. Let's figure out what's the greatest common factor of 12, 6, and 15. And there's a couple of ways you could do it. You could do it by looking at a prime factorization. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | The first thing we can look at is let's look at the coefficients. Let's figure out what's the greatest common factor of 12, 6, and 15. And there's a couple of ways you could do it. You could do it by looking at a prime factorization. You could say, all right, well 12 is two times six, which is two times three. That's the prime factorization of 12. Prime factorization of six is just two times three. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | You could do it by looking at a prime factorization. You could say, all right, well 12 is two times six, which is two times three. That's the prime factorization of 12. Prime factorization of six is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there. Three is divisible into all of them. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | Prime factorization of six is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there. Three is divisible into all of them. And that's it, because we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five, because five isn't divisible into 12 or six. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | Three is divisible into all of them. And that's it, because we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five, because five isn't divisible into 12 or six. So the greatest common factor is going to be three. Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | We can't say a three and a five, because five isn't divisible into 12 or six. So the greatest common factor is going to be three. Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. Six, you could have said, let's see, that could be one times six, or two times three. So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. Six, you could have said, let's see, that could be one times six, or two times three. So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here. We could write, we could write that, excuse me, I wanna see the original thing. |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | We're told a roller coaster has C cars, each containing 20 seats, and it completes R rides a day. Assuming that no one can ride it more than once a day, the maximum number of people that can ride the roller coaster in a single day is P. Write an equation that relates P, C, and R. Pause this video and see if you can do that. All right, before I even look at the variables, I'm just gonna try to think it out in plain language. So what we wanna think about is what is the max number of people per day? People per day. And so that's going to be equal to the number of cars in our roller coaster. So number of cars times the maximum number of people per car times the maximum number of people per car times the maximum number per car. |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | So what we wanna think about is what is the max number of people per day? People per day. And so that's going to be equal to the number of cars in our roller coaster. So number of cars times the maximum number of people per car times the maximum number of people per car times the maximum number per car. So this would just tell you the maximum number of people per ride. So then we have to multiply it times the number of rides per day. So times, we'll do this in a new color, times number of rides per day. |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | So number of cars times the maximum number of people per car times the maximum number of people per car times the maximum number per car. So this would just tell you the maximum number of people per ride. So then we have to multiply it times the number of rides per day. So times, we'll do this in a new color, times number of rides per day. So what are each of these things? They would have either given us numbers or variables for each of them. The max number of people per day, that's what we're trying to set on one side of the equation. |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | So times, we'll do this in a new color, times number of rides per day. So what are each of these things? They would have either given us numbers or variables for each of them. The max number of people per day, that's what we're trying to set on one side of the equation. That is this variable P right over here. So we'll say capital P is equal to what's the number of cars per coaster, I guess you could say. Let me write it this way, per coaster, per roller coaster. |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | The max number of people per day, that's what we're trying to set on one side of the equation. That is this variable P right over here. So we'll say capital P is equal to what's the number of cars per coaster, I guess you could say. Let me write it this way, per coaster, per roller coaster. So they give us that right over here. A roller coaster has C cars. So that's going to be this variable here in orange or this part of it, that C. Now what's the maximum number of people per car? |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | Let me write it this way, per coaster, per roller coaster. So they give us that right over here. A roller coaster has C cars. So that's going to be this variable here in orange or this part of it, that C. Now what's the maximum number of people per car? Well, they say each containing 20 seats. So I'd multiply that times 20 for this part. And then I want to multiply that times the number of rides per day for the entire roller coaster. |
Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3 | So that's going to be this variable here in orange or this part of it, that C. Now what's the maximum number of people per car? Well, they say each containing 20 seats. So I'd multiply that times 20 for this part. And then I want to multiply that times the number of rides per day for the entire roller coaster. So that's going to be times R. And we're done, we could rearrange this a little bit. We could write this as P is equal to 20 times CR. 20 times CR. |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | And as we'll see it a little, it may feel a little magical at first, but in future videos we will prove it and we'll see, well, like many things in mathematics, when you actually think it through, maybe it's not so much magic. So what is the polynomial remainder theorem? Well it tells us that if we start with some polynomial, f of x, so this right over here is a polynomial, polynomial, and we divide it, we divide, divide by x minus a, then the remainder, then the remainder from that, essentially, polynomial long division is going to be f of a. It is going to be, it is going to be f of a. f of a, I know this might seem a little bit abstract right now, I'm talking about f of x's and x minus a's, let's make it a little bit more concrete. So let's say that f of x, f of x is equal to, I'm just gonna make up a, let's say a second degree polynomial. This would be true for any polynomial though. So three x squared minus four x plus seven, and let's say that a is, I don't know, a is one. |