- Positive and negative intervals of polynomials
- Positive & negative intervals of polynomials
- Multiplicity of zeros of polynomials
- Zeros of polynomials (multiplicity)
- Zeros of polynomials (multiplicity)
- Zeros of polynomials & their graphs
- Positive & negative intervals of polynomials
Given the graph of a polynomial and looking at its x-intercepts, we can determine the factors the polynomial must have. Additionally, we can determine whether those factors are raised to an odd power or to an even power (this is called the multiplicity of the factors).
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- What is the real life implement of such types of problems.
Why do we need to study these problems.
My curious mind always thinks about it.(5 votes)
- A curious mind is key to unlock the meaning of mathematics. Of course, I cannot assume what you want to be, that's your decision.
Polynomial roots can be found in the real life through data analysis of economics, sciences such as chemistry and physics, calculating costs of mortage balance, and much more. You have lots to find, as I cannot list all the ways that polynomial roots--rather, polynomials in general--in one post. Hopefully that helps !(12 votes)
- How do you find the multiplicity and zeros of a polynomial?(4 votes)
- Wherever the polynomial touches the X axes a Zero is located there. Multiplicity is the type of polynomial that touches the x axis(1 vote)
- Is very confusing because in the video they explain they explain when a number is odd doesnt have a exponent but in the exercise they show as it should have(3 votes)
- if there is sign change, the exponent of that expression of the root would be raised to an odd power it doesn't necessarily have to be raised to the power of 1, it could be 3 or 7 or any odd number.
root of -3, sign change around it : (x+3)^odd exponent
if there is no sign change, expression will be raised to an even power
root of 2, no sign change : (x+2)^even exponent(1 vote)
- What does Sal mean when he says the sign changes?(2 votes)
- By sign change, he mans that the Y value changes from positive to negative or vice versa. For example, if you just had (x+4), it would change from positive to negative or negative to positive (since it is an odd numbered power) but (x+4)^2 would not "sign change" because the power is even(2 votes)
- the consecutive zero's rule doesn't seem all the consecutive. One question referenced how -5<x<2/7, but it was referred to as a consecutive zero which I don't understand because it passes the 0.(1 vote)
- A zero of a function is when the y-value equals zero: that is, when the function crosses the x-axis. I am guessing that for this problem when (x=-5, y=0), and when (x=2/7, y=0). If there is another time between them that it crosses the x-axis, then they are not consecutive, however; if you just mean that it passed 0 in the x direction, then they could still be consecutive. I hope that this helps.(3 votes)
- What happens to the point on the graph where x=0? Do we exclude any sign change here?(1 vote)
- No, sign changes only occur when the y-value switches from positive to negative or vice versa. You may have a sign change if y=0, x can =0 as well but y must.(1 vote)
- What do you mean by sign change tho(1 vote)
- "Sign Change" means the signs of the terms switch from positive to negative or negative to positive. For example, in f(x)=-3x^2 +2x -4, there are 2 sign changes: -3x^2 to +2x, then +2x to -4. Hope this helps.(1 vote)
- None of this works in the test - literally the first example had an exponent of 2 and chnaged from positive to negative?! How is this possible based on what you say in the video?(1 vote)
- I don't know what exact problem you're talking about, but if you find a mistake you can report it. If you're still confused about it, you can bring it to a free online tutoring session as schoolhouse.world
Hope this helps!(1 vote)
- Would y = c (parallel to x-axis) have any zero (real or complex)?
(in our textbook) it is given that a polynomial must have at least 1 zero, (not necessarily real nor distinct), is this statement false?)(1 vote)
- Any nonzero constant polynomial f(x)=c has no roots. No matter what x you input, you get c, not 0.
This is inkeeping with the fundamental theorem of algebra. Constants are degree-0 polynomials, and they have 0 roots.
Edit: Your textbook is not quite correct, since degree-0 polynomials have no roots. However, a polynomial of degree at least 1 will indeed always have a root.(1 vote)
- 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.