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Factoring higher-degree polynomials

Sal factors p(x)=2x^5+x^4-2x-1 as (2x+1)(x^4-1) using grouping. Then he further factors (x^4-1) as (x^2+1)(x+1)(x-1) using the special product form of difference of squares. Created by Sal Khan.

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  • duskpin ultimate style avatar for user Roshni
    At , it says "x^2=-1." That means that x = imaginary number i. How could we graph i on a coordinate plane?
    (14 votes)
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    • leaf green style avatar for user Steven
      When graphing complex numbers, there is an axis for the real portion and another axis for the imaginary portion. If I remember correctly, the x-axis will be your real axis and the y-axis will be the imaginary axis. For example, if I have the complex number 1 + 2i, the point will look like the point (1, 2) on a set of real axes.
      (18 votes)
  • piceratops tree style avatar for user popbampop
    If a real zero is a place where the function intersects the x-axis then does the function sal solved intersect the x axis 3 times?
    (11 votes)
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  • duskpin ultimate style avatar for user grigor21
    The Fundamental Theorem of Algebra tells that any polynomial of degree n has exactly n roots, but Sal in the video found 3 roots for the polynomial of 5 degree. Please explain how is this possible?
    (5 votes)
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  • blobby green style avatar for user jamie_chu78
    Isn't there a set a rules one could follow when factoring x^n power? It seems that the way in which you factor x^2 is completely distinct from how you would factor x^3, or x^4 and so on. There seems to be no general order to factoring exponents, which is what I really do not gel with...
    (6 votes)
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    • male robot hal style avatar for user Ryan Stones
      Unfortunately, there is NOT a set a rules for factoring when you get to higher powers. Grouping can sometimes work--and when it does, it works well--but that method requires a pattern in the numbers. What mathematicians do to find roots in the more difficult cases is to a) use synthetic division with the rational root theorem to guess and check possibilities for rational roots or b) use graphing to identify the real roots (which might only lead to an approximation, not an exact root).

      There are (much more difficult) formulas like the quadratic formula for degree x^3 and x^4, but it's actually a deep mathematical theorem (and fascinating historical story) that there can be no formula for degree x^5 polynomials or higher. So unfortunately, even though the Fundamental Theorem guarantees roots, practically speaking, it might be impossible to identify them sometimes.
      (4 votes)
  • aqualine seed style avatar for user Madeline
    when he further factored (x^4-1) to (x^2-1) and (x^2+1) why is that necessary?
    (3 votes)
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  • duskpin ultimate style avatar for user Tara Andreoni
    h(x)=-x^3-5x^2 how do I find the zeros of the function
    (2 votes)
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  • piceratops ultimate style avatar for user Julia Cecchetti
    If you get an imaginary root, how would you graph that on the coordinate system?
    (2 votes)
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    • male robot hal style avatar for user Aris Martinian
      You can't. Imaginary numbers can only be graphed on a complex plane. Typically you would see something like i replacing the y-direction. You could draw another coordinate system to suit your needs or even draw a z-direction and plot it in 3 dimensions. Then you could plot i and -i on the new imaginary direction.
      (2 votes)
  • blobby green style avatar for user haziqahsahrizal
    How to factor x^4 + 2x^3 + 2x^2 + 2x + 1
    (2 votes)
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    • leaf green style avatar for user kubleeka
      Guess-and-checking a few simple numbers, I found that i is a root. Because this polynomial has real coefficients, that means that the complex conjugate -i is also a root. So we can factor out
      (x+i)(x-i)=x²+1 with synthetic division.

      This gives us (x²+2x+1)(x²+1). Now we can use the quadratic formula to find the roots of x²+2x+1.

      We get [-2±√(4-4)]/2= -2/2= -1. So -1 is a double root, so this factors as (x+1)²(x²+1).
      (2 votes)
  • aqualine ultimate style avatar for user Ahmadshaheerhakimi
    One of the zeroes that you got was -1/2 doesn't that mean you have to go 1 left and 2 up on the graph
    (1 vote)
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    • leafers sapling style avatar for user Cata Faerman
      Nopee. That's only when you are using the formula of a slope, because the formula tells you m=y2-y1/x2-x1 (hence, 1 down, 2 to the right).

      That the "zero" is -1/2 means that when X= -1/2, your Y-coordinate is going to be "0". Therefore, the point is going to be in the coordinate (-1/2, 0).

      Remember that a "zero" or "root" is a point where the graph touches the X-axis.
      (4 votes)
  • blobby green style avatar for user PadamG90
    10x^5-5x^4-4x^3-3x^2-2x-1=0
    (2 votes)
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Video transcript

Plot the real zeroes of the given polynomial on the graph below. And they give us p of x is equal to 2 x to the 5th plus x to the 4th minus 2x minus 1. And when they say plot it, they give us this little widget here. Where if we click at any point on this, we get our point. And we get as many points as we would like. And we can drag these points around. Or if we don't want these points anymore, we can just dump them in this little trash bin at the bottom right. So let's think about what the zeroes of this polynomial actually are. And to do that I'll take my scratch pad out. And this is a little daunting at first. This is a 5th degree polynomial here. Factoring 5th degree polynomials is really something of an art. You're really going to have to sit and look for patterns. If they're actually expecting you to find the zeroes here without the help of a computer, without the help of a calculator, then there must be some type of pattern that you can pick out here. So let me just rewrite p of x. So p of x is equal to 2 x to the 5th plus x to the 4th minus 2x minus 1. And one way that's typically seen when you're trying to factor this type of polynomial is to try to essentially undo the distributive property a few times. And if you want to relate it to techniques for factoring quadratics, it's essentially factoring by grouping. So for example, you see a 2x minus 1, or something that looks like a 2x minus 1 right over here. And over here you have a 2x to the 5th plus x to the 4th. So you have a 2x of a higher degree term plus a 1 x of a one degree lower. So there seems to be some type of a pattern. 2 times x of a higher degree-- this is a first degree term-- minus 1 times-- you could do this as x to the 0-- of a lower degree term. And so let's think about it a little bit. What happens if we essentially try to group these two terms, and we group these two terms right over here. And we try to factor out anything to essentially clean it up a little bit, to see if we can make sense of it. Well these two terms, their greatest common factor is x to the 4th. We could write this as x to the 4th times 2x plus 1. And this should get us excited because this looks pretty close to that, especially if we were to factor out a negative 1 here. So we could factor out a negative 1. And then this is going to be 2x plus 1. And that's exciting because now we can factor out a 2 x plus 1 from each of these terms. So you have a 2x plus 1. We're gonna factor both of these out, to get 2x plus 1-- which we just factored out. And if you factored it out of this term right over here, you're left with x to the 4th. And if you factor it out this term, you're left with just the minus 1. Minus 1. And now this is exciting because 2x plus 1, this is pretty easy to figure out when does this thing equal 0. And we'll do that in a little bit. And this is pretty easy to factor. This is a difference of squares. This right over here can be rewritten as x squared plus 1 times x squared minus 1. And of course we still have this 2x plus 1 out front. 2x plus 1. And once again we have another difference of squares. We have another difference of squares right over here. That's the same thing as x plus 1 times x minus 1. And let me just write all the other parts of this expression. x squared plus 1. And you have 2x plus 1. 2x plus 1. And I think I factored p of x about as much as could be reasonably expected. So p of x is equal to all of this business. And remember, the whole reason why I wanted to factor it is I wanted to figure out when does this thing equal 0? So if p of x can be expressed as the product of a bunch of these expressions, it's going to be 0 whenever at least one of these expressions is equal to 0. If any of these is equal to 0 then that's just going to make this whole expression equal to 0. So when does 2x plus 1 equal 0? So 2x plus 1 is equal to zero. Well you could probably do this in your head, or we could do it systematically as well. Subtract 1 from both sides, you get 2x equals negative 1. Divide both sides by 2, you get x is equal to negative 1/2. So when x equals negative 1/2-- or one way to think about it, p of negative 1/2 is 0. So p of negative 1/2 is 0. So this right over here is a point on the graph, and it is one of the real zeroes. Now we could try to solve this. x squared plus 1 equals 0, and I'll just write it down just to show you. If we try to isolate the x term on the left-- subtract 1 from both sides-- you get x squared is equal to negative 1. Now if we were to start thinking about imaginary numbers, we could think about what x could be. But they want us to find the real zeroes. The real zeroes. So there's no real number where that number squared is equal to negative 1. So we're not going to get any real zeroes by setting this thing equal to 0. There's no real number x where x squared plus 1 is going to be equal to 0. Now let's think about when x plus 1 could be equal to 0. We'll subtract 1 from both sides. You get x is equal to negative 1. So p of negative 1 is going to be 0. So that's another one of our zeroes right there. And then finally let's think about when x minus 1 is equal to 0. We'll add one to both sides. X is equal to 1. So we have another real 0 right over there. And so we could plot them. So it's negative 1, negative 1/2, and 1. So it's negative 1, negative 1/2 and 1. And we can check our answer, and we got it right. Now one thing that might be bugging you, is like hey Sal, you just happened to group this in exactly the right way. What if I try to group it in a different way? What if I tried to-- Well actually let's try to do that. That could be interesting. Just to show you that this isn't voodoo. And actually there's several ways to get there. There's several ways to get there. So what if instead of writing it like this, where you're writing it kind of in the highest degree term and the next highest degree and so on and so forth, you were to write it like this: p of x is equal to 2 x to the 5th minus 2x plus x to the 4th minus 1. Well actually, even this way you could do a fairly interesting grouping. If you grouped these two together, you see that they have the common factor 2 x. You factor 2x out, you get 2x times x to the 4th minus 1. And I think you see what's going on. And then this can be rewritten as plus 1 times x to the 4th minus 1. Minus 1. And now you can factor out an x to the 4th minus 1. And you're left with-- I'll do this in a neutral color-- x to the fourth minus 1 times 2 x plus 1, which is much easier to factor now. A difference of squares. Exactly what we did the last time around. So there are several ways that you could have reasonably grouped this, and reasonably undone the distributive property. But I do admit it is something of an art. You really just have to play around in see, well let's group the first two terms, let's see if there's a common factor here. Let's group the second two terms, let's see if there's a common factor here. Hey, once we factor out those common factors, it looks like both of these two terms have this common expression as a factor. And then you could start to factor that out.