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Polynomial factors and graphs — Harder example

Watch Sal work through a harder Polynomial factors and graphs problem.

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  • piceratops seed style avatar for user ganterma
    find a harder example. This is way too easy when all you have to do is count the x intercepts and how many times the graph touches the xaxis
    (83 votes)
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  • blobby green style avatar for user Joudi
    Just look at the exponents if the exponents are even the graph lines go to the same direction
    Since the graph has both arrows down the exponent SHOULD be EVEN
    (35 votes)
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  • mr pants teal style avatar for user Marco Royce
    what is the difference between distinct zero and double zero?Why would the graph bounce back?What is the official definition of double zero?Thanks.
    (14 votes)
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    • aqualine tree style avatar for user Ravengal101
      A double zero results from a function having a repeated root, for example: roots derived from factors of the form (x-a)^2. We already know that roots occur where the graph touches/cuts the x axis, so if a factor is of some squared form then the corresponding y values of the function would be positive. At the point of the root, the graph doesn't cross the x axis (because a quadratic function governs that portion of the graph) but instead bounces back from the x axis. Get it?
      (17 votes)
  • blobby green style avatar for user hamzakhan2479777
    Can’t i just apply the rule that if x power’s even then both negative and positive curves goes in the same direction and if it is negatives then doesn’t
    (17 votes)
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  • blobby green style avatar for user Devika Bhatnagar
    What's the difference between distinct and double zeros?
    (10 votes)
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  • blobby green style avatar for user Magduuuh
    one can also use the odd and even aspect right? Like X raised to 1 which is odd hence shud face diff directions and X raised to 2 which is even facing same direction downwards.
    (11 votes)
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  • duskpin ultimate style avatar for user Kalen
    I don't get why A does not work. Something about end behavior.
    (4 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      You can explain why A doesn't work 2 ways. You could talk about end behavior: You should know that the end behaviour of polynomials is determined by if their degree is an odd or even power: if odd, then the ends go in opposite directions and if even, the ends goin the same direction. Here, our ends are going in the same direction so we need to pick something with an even degree. Adding up all the x's you see, for A) we get a degree of 3 and for B) we get 4. This allows us to rule out A).
      Another way you can approach this is by thinking about multiplicity of roots, or how often the root is repeated (or when the polynomial is shown in root form like here, what the exponent on each root is). Roots with multiplicity 1 have the graph passing through them, roots with even multiplicity have the graph touch the root and "bounce" backwards, and roots with odd multiplicity greater than 1 have the graph do a little "squiggle", where it becomes horizontal at the root but then eventually crosses it. Here, at x=0 the graph goes down to 0 but then bounces back, so we need an even multiplicity for that root, which A) does not have and B) does have.
      (13 votes)
  • female robot grace style avatar for user Meher Pannu
    Can you find a harder question like from a practice test or something? The hard questions aren't supposed to be this easy.
    (9 votes)
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  • blobby green style avatar for user muhammadahmadkhan2812
    https://www.khanacademy.org/math/algebra-home/alg-polynomials

    Watch these if you have problems in understanding this
    (8 votes)
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  • blobby green style avatar for user Mohammed Sulaiman Javed
    He’s not going to answer to your questions.... don’t ask him
    (7 votes)
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Video transcript

- [Instructor] We're asked which of the following could be the equation of the graph above. So pause this video and see if you can work through this before we do it together. All right, so first of all, let's just think about where the graph intersects the X axis. We can see it does that at X equals negative four. It does that at X equals zero. And it does that at X equals positive three. So we would expect to see an expression that's equal to zero at these three places. So first of all, X equals zero. You could see if X equals zero. All of these will become zero, 'cause zero times anything else is zero. So they all meet that one. See, X equals negative four. In order for it to become zero at X equals negative four, we'd want to see an X plus four when we factor things out. We see an X plus four here, we see an X plus four here. These second two, they have X minus four. So we don't like those. So we can immediately rule those out. And then for X equals three to make the entire expression equal to zero, you'd need to have an X minus three. And we can see we have an X minus three and an X minus three. So now we have to decide what makes more sense to have a negative X out here or a negative X squared. Well, let's think about what would happen in either case as X increases. As X increases, as X becomes larger and larger positive values, you take the negative of that, and it makes sense in this A choice that the value would get more and more negative. That's also the case in choice B. As X increases, you square it. You'll get even a more positive value. And then you take the negative of that, it should go down. So both choice A and B is consistent with this behavior of as X becomes more positive, Y is becoming more and more negative. Now let's look on the other hand. What about when X becomes more and more negative? Well, when X becomes more and more negative in choice A, when you take the negative of a negative you would expect this part to become more and more positive. So you would expect for choice A, that actually the Y values would increase as X becomes more and more negative, which clearly isn't happening. So we can rule that one out, as well. But let's just verify that B works. So when X becomes more and more negative, you take the square of that. So you're going to get a positive value, but then you take the negative of that. So it's going to become, so Y will keep decreasing and becoming more and more negative, which is exactly what we see in this graph. So we're liking choice B.