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- All right, now let's work through this together. And we can see that all of the choices are expressed as a polynomial in factored form. And factored form is useful when we're thinking about the roots of a polynomial, the x-values that make that polynomial equal to zero. The roots are also evident when we look at this graph here. We have a root at x equals -4, a root at x equals -1.5, or -3/2, and a root at x is equal to 1. 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 -4. In order for this polynomial to be zero when x is equal to -4, that means that x + 4 must be a factor, or some multiple, or some constant times x + 4, must be a factor of our polynomial. Now we can see in the choices that we have a bunch of x + 4s, but they have different exponents on them. The first one has a 2 as an exponent, it's being squared, while the others have a 1 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 -4," 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, 1 is an odd number and 2 isn't. So if you just have a straight up x + 4, you would have a sign change around x equals -4. So I can rule out this first choice, these other three choices are still looking good based on just the x + 4 factor. Now lets move on to the next factor right over here, so, or the next root. The next root is at x is equal to -3/2, and so one way to think about it is you could have a factor that looks like x + 3/2, or this times some constant. Now, when we look at the choices, or the remaining choices, we don't see x + 3/2, but we do see something that involves a 2 and a 3, and so one way to think about it is just, "Hey, if I just multiply this by the constant 2, that would get us 2x + 3," 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 -3/2, so we would expect an odd exponent there. And you can see out of the choices, only two of them have an exponent of 1, 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. Ah, we'll do this in an orange color. We have a root at x equals 1, so we would expect x - 1, or this multiplied by some constant, to be one of the factors. And what's interesting here is we don't see a sign change around x equals 1, so we would expect an even exponent. And so, out of the remaining choices, we see an x - 1 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 look at choice D, where this is to the first power, we would expect a sign change around x is equal to 1, 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 going to do the same idea. Let's go to this first root right over here. We have a root at x equals -3, so we would expect some multiple of x + 3 to be one of the factors. We also have a sign change around x equals -3, so we would expect an odd multiplicity and we would expect an odd exponent on the x + 3 factor. When we look at all the choices, C and D have an even exponent, so if we had x + 3 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 2, so we would expect x - 2 to be one of the factors, or a multiple of this, and because we don't have a sign change around x equals 2, the graph just touches the x-axis and goes back to where it was coming from, we would expect an even exponent here. And so when we look at the choices around the x -2 factor, we see only one of them has an even exponent, so I am liking Choice B, and we are done.