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Finding zeros of polynomials (example 2)

Sal finds all the zeros (which is the same as the roots) of p(x)=(3x⁴-8x³+15x-40)(3x-8)².

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  • blobby green style avatar for user Michael Gomez
    at , why did he not take into account the ^3 on 3x-8 when he was trying to figure out why it was equal to zero
    (6 votes)
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  • piceratops ultimate style avatar for user James Warburton
    At why is the answer not -1.71 i? I thought if you take the the root of a negative number you have to put i after the answer to specify that it is imaginary.
    (3 votes)
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  • winston baby style avatar for user Samuel Lee
    I tried doing the "Practice:Zeros of polynomials challenge problems", but I could not understand anything whatsoever. Can someone explain to me how to solve those problems?
    (10 votes)
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    • blobby yellow style avatar for user SARIN V S
      Take any polynomial first of all, factorize all the polynomial to a form like this, say [2x-x][10x-8x], and now equate this = 0 to earn zeroes of the polynomial. To know the zero of the polynomial either any one of the brackets should be equal to zero. Now in the first bracket, it turns out to be 2x-x=x so x = 0... In the second bracket 10x-8x=2x and if 2x = 0 then x= 0/2=0 so it turned out to be that 0 and 0 are the "zeros of the polynomial". This means 0 is the "zero" of this polynomial [2x-x][10x-8x]...

      Always 0 may not be the "zero of the polynomial". It can be any number.
      (2 votes)
  • male robot hal style avatar for user Hanuman
    Just to make sure I understand this:
    To find the zero of a polynomial, you have to factor the polynomial until you vet the last expressions. In this case, the polynomial simplified to (3x-8)^3 x (x^3+5). So we just solve (3x-8)^3 for 0 and then (x^3+5) for 0. To get 8/3 and sqrt3. -5 right?. If you enter these x values in the function, it will intercept the x axis, right? Also how would this be useful in real life?
    (3 votes)
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    • duskpin ultimate style avatar for user J E
      Yes that is correct, (3x-8)^3 * (x^3 +5) has the zeros x=8/3 and x= -cuberoot(5).
      How is this useful in real life? Well, imagine you're throwing a ball and you want to know how far away from you it will hit the ground. If we think of the ground as the x-axis and yourself as the y-axis (your position is at x=0) then using your equation (and there are ways to find the equation of the ball's trajectory) you can find the zeros of the parabolic function and use them to see how far away from you the ball lands.
      Hope this helps!
      (7 votes)
  • male robot hal style avatar for user Ludo Constant
    how would we find the four remaining non-real complex roots?
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      This polynomial has no complex roots.
      Yes, the polynomial should have 6 roots. However, there reason we have only 2 roots is because there are duplicates.
      (3x-8)^3=0 means there are 3 roots that are exactly the same.
      If you write this without the exponent, you see the 3 factors
      (3x-8)(3x-8)(3x-8)=0
      Use the zero produce rule, separate & solve and you get 3 identical values:
      3x-8 = 0 creates 8/3
      3x-8 = 0 creates 8/3
      3x-8 = 0 creates 8/3
      This is called multiplicity-- Multiple roots that are the same.
      The same thing is happening with x^3 = -5. This represents 3 matching roots.
      So, all 6 roots are accounted for, there are just 4 values that are duplicates.
      Hope this helps.
      (1 vote)
  • blobby green style avatar for user Fred Haynes
    In the Practice Problems - Zeros of polynomials challenge problems, I can't figure out how to work any of those problems. Below is one example, all of the problems are similar.

    The position, p, of a particle is expressed in terms of time as:

    p=(a-t)(t-b)(b-t) where a>b


    I can't figure out how to work any of those challenge problems. Any help would be greatly appreciated. This is the only section that totally makes no sense to me.

    Thanks in advance.
    (2 votes)
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  • aqualine ultimate style avatar for user Nadeem Hussein
    I understand finding the roots of a polynomial given that at least one of the roots are rational, but how would I find the EXACT roots of a polynomial (mainly a trinominal with four terms) given that all the roots are either irrational or imaginary?

    Example: x^3 -2x^2 -4x -6
    Grouping doesn't work here
    I've tried using the rational roots theorem, then plugging each possible root into the equation, none of them are factors
    I don't understand how to find the exact roots of the polynomial given that there are no rational roots.
    (1 vote)
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    • leaf green style avatar for user kubleeka
      Finding exact roots of a polynomial is, in general, a genuinely hard problem. After polynomials came into popular use, it was about 300 years before we had a decent set of tools for finding roots, with a branch of math called Galois theory.

      As it happens, there is an explicit formula for finding the roots of degree-3 polynomials, and another, even more complicated one for degree-4. But there is not, and cannot be, such a formula for equations of degree 5 and up.

      The way to find roots of 3rd-degree equations is described here:
      https://qr.ae/TUhUjK
      (3 votes)
  • primosaur ultimate style avatar for user scragglegrackle
    How are the other 2 complex roots found in this equation?
    (2 votes)
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  • leaf green style avatar for user krisgoku2
    how is 8/3 equal to zero?
    (2 votes)
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  • starky sapling style avatar for user Genghis Khan
    If, for example, we had a 3rd degree polynomial with a double root, (and its other root was a real), would it intercept the x-axis 2 or 3 times?
    In one of the practice hints it says that in this case, the polynomial would only have 2 "distinct roots". Does that mean the same thing as times it intercepts the x axis, or are they different?
    (1 vote)
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    • leaf green style avatar for user kubleeka
      It would cross the x-axis at one point and touch it without crossing at the point with the double root.

      The polynomial (x-2)²(x-3) has 3 real roots, but two of those roots are 2. So if we listed all of the numbers that are roots of this polynomial, we would write 2, 3. So we say it has two distinct roots, since there are only two numbers in the list.
      (3 votes)

Video transcript

- So I have the polynomial p of x here, and p of x is being expressed as a fourth degree polynomial times three x minus eight squared, so this would actually give you some, this would give you nine x squared and a bunch of other stuff, and then you multiply that times this. It would actually give you a sixth degree polynomial all in all, but our goal is to find the x values where that makes p of x equal to zero, or another way find the roots or the zeros of this polynomial, and in particular we're going to focus on the real zeros, the real roots of this polynomial, and like always I encourage you give a go at it, and then we'll do it together. Alright, let's tackle this. So, the way I want to solve p of x is equal to zero. I want to solve p of x is equal to zero, and figure out what, and when I say solve it I want to say well what x values will make the polynomial equal to zero. So I just need to set this right hand side to be equal to zero and then solve for x, and the best way that I can think of doing that is by factoring this out as much as I can, and if I can rewrite it as a product of a bunch of expressions equaling zero, well a product of a bunch of things equaling zero, you can make it equal zero by any one of them equaling zero, and so let's do that. So three x minus eight squared, this is already factored quite nicely, let's see if we can factor if we can factor all of this business in white, and the way I will tackle it to see if I can factor by grouping. So let me group together these first two terms, and then let me group together these second two terms, and essentially factoring by grouping is doing the distributive property in reverse twice. So from these first two terms, I could factor out, let's see what could I factor out? I could factor out a let me see, I'll just factor out an x to the third power. So I get x to the third power times three x minus eight, interesting we have a three x minus eight over there as well Now these second to two terms, I could factor out a five, so this is going to be plus five times x, sorry, times three x. Three x minus eight, very interesting, and of course I have these parentheses around all of that, and then I have three x minus 8 squared. This three x minus eight is showing up a lot, and so and of course this is going to be equal to zero. So we're gonna be equal to zero, and now I can factor out a three x minus eight over here, I could factor that three x minus eight out, and I'm going to get three x minus eight times times x to the third power, x to the third power plus five. Once again I just factor out a three x minus eight plus five, close the parentheses, and then times three x minus eight squared is all going to be equal to zero, is all equal zero. Now if what I just did looks a little like voodoo, just realize I have two terms, both of them are multiples of three x minus eight. I just factored out, I just factored out the three x minus eight. I did distributive property in reverse, so I factored it out, and what you're left with this term you just look for the next of third, and in this term you're just left with a plus five. Now three x minus eight times x to the third plus five times three x minus eight squared. Well I could just rewrite this as three x minus eight to the third power times x to the third plus five, so let me do that. So I can just rewrite this as three x minus eight to the third power, that's that times that, and then times times, do this in a nicer color, times x to the third plus five is equal to zero, is equal to zero, and now in order to get this to be equal zero, either the three x either this thing is going to be equal to zero, or this thing over here is going to be equal to zero. So let's first think about three minus eight to the third power equaling zero. So I can write this is as three x minus eight to the third power is equal to zero or x to the third plus five is equal to zero. So to make three x minus eight to the third power equal zero well that means three x minus eight is going to be equal to zero or that three x is equal to eight, divide both sides by three, x is equal to eight thirds. So that's one way to make this polynomial equal zero, x is equal to eight thirds, in fact just this right over there will become zero, zero times anything is zero. So this is a zero of our polynomial, and let's see, so x to the third, so we could say, if we subtract five from both sides, we have x to the third is equal to the is equal to negative five, and so if we take both to the one third power we could say x is equal to cube root of negative five. Now at first you might say, "Wait, can I take the square root of a negative number?" and I would say, "Of course you can!" The cube root of a negative one is negative one. The cube root of negative eight is negative two. In fact you could, even if we're dealing with reals. This is going to be a negative number. This is not going to be an imaginary number right over here, and so these are, these are the two zeros of the polynomial. There's gonna be negative, I think, negative one point something, I'm sure we could figure out it, figure out it exactly. So let's raise, so let's raise, five to the open parentheses, one divided by three close parentheses, is equal to, so that's five to the one third, so negative five to the one third power is gonna be negative one point seven one approximately. So this is approximately equal negative one point seven one. So we have two real roots, two real roots to this polynomial, or two zeros two real zeros for this polynomial, and so those are going to be the two places where we intercept the x-axis. The two x values for which where the two places where we intercept the x-axis, is the easiest way to say it.