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Factoring higher degree polynomials

Factoring a partially factored polynomial and factoring a third degree polynomial by grouping.

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  • aqualine ultimate style avatar for user Angel C.
    What are some common real world applications for this?
    (21 votes)
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    • winston default style avatar for user J
      You may need to use factoring often if you have a real world job. If you decide to become an economist, statistician, engineer, mathematician, or any kind of physical scientist. You would commonly use factoring. For example, data containing complex algebraic fractions can look extremely difficult. You would need to factor in order to simplify and make use of the data in an easier way. Basically, it will make your life more simple though it seems annoying right now. Hope this helps!
      (59 votes)
  • leaf red style avatar for user jerimiah.moore
    So I have watched video after video and I also took notes.


    I even took some hints I still cannot get this please help.
    (14 votes)
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    • duskpin ultimate style avatar for user Rachael
      The first example is set up like this:
      (6x^2 + 9x)(x^2 - 4x +4).
      What Sal did was take the GCF out of each set of parentheses. As you may have seen in previous videos in this unit, the way to find the GCF from the left set of parentheses is to find the GCF of the coefficients. In this case, the GCF of 6 and 9 is 3. The next step to find the GCF of the full terms is to look at the variables. There is an x^2 and an x on its own. Let's list out the factors:
      x^2 factors are 1 and x^2, and x and x since those numbers multiply to get to x^2.
      x factors are x and 1.
      We can see by looking at the factors that the greatest factor is x (assuming x is greater than one, but we aren't going to worry about that because we don't have to solve for x in this problem). Since the GCF of the variables is x and the GCF of the coefficients is 3, we multiply them together to get 3x.
      Now that we have our GCF for the left set of parentheses, we can divide everything in the left set by our GCF, and bring the GCF out of the parentheses. This is better illustrated in the video Taking common factor from binomial, under Taking common factors earlier in this unit, so I am going to skip explaining that part.
      (6x^2 + 9x)/3x is (2x + 3). Now we bring the 3x to the outside of the parentheses to get 3x(2x + 3).
      That's one half of the equation. The other we can tell just by looking that it is a perfect square, so we split it apart as shown in the first unit called Polynomial Arithmetic, with the video Polynomial special products: perfect square.
      Splitting (x^2 - 4x + 4) into its square roots results in this:
      (x - 2)(x - 2).
      The next step is to put all of that together. This gets us
      3x(2x + 3)(x - 2)(x - 2) Since you can no longer factor this equation, it is in simplest form. That means we just leave it like that.

      The second example is a little different:
      x^3 - 4x^2 + 6x - 24.
      The easiest way to solve this is to factor by grouping. To do that, you put parentheses around the first two terms and the second two terms.
      (x^3 - 4x^2) + (6x - 24). Now we take out the GCF from both equations and move it to the outside of the parentheses.
      x^2(x - 4) + 6(x - 4). As you can see, the sets of numbers inside the parentheses are the same. This means that we can take the numbers outside the parentheses and put them in their own set.
      (x^2 + 6)(x + 4). When you multiply that out, you get x^3 - 4x^2 + 6x + 24. That means that this is as simplified as you can get your equation. Also, it means you just did all of that math to get a circle (start in one place, end in the same).
      (23 votes)
  • leafers sapling style avatar for user Draven Brown
    I'm still confused on this still after watching an hour.
    (18 votes)
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  • duskpin ultimate style avatar for user LarissaAnne
    At he mentions there are other videos that he assumes we've already seen. Which videos are they?
    (7 votes)
  • aqualine ultimate style avatar for user Ŧг๏รՇ ๏ฬɭ
    im confused on how he does it is there more then one way to factor this same problem?
    (3 votes)
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    • eggleston yellow style avatar for user Simon
      Well, it depends. If there is 4 terms, 2 quadratics multiplied, 3 terms, it would need to be factored differently. Try doing some of the practice problems for factoring higher degree polynomials', and you'll understand what I'm talking about.
      (1 vote)
  • duskpin sapling style avatar for user Nóra McGowan
    Is there a way to get like a crash course on factoring polynomials? I really need to understand it, and fast.
    (3 votes)
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  • cacteye green style avatar for user Peter Lai
    How do you factor expressions with a difference of squares pattern? Please elaborate as it only shows this type of problems in practice.
    (2 votes)
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  • cacteye green style avatar for user Peter Lai
    at , is there an easy way to come up with two numbers whose is four and whose sum is negative four? He seems to figure it out very quickly so is there a video that I so happened to missed?
    (2 votes)
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  • female robot ada style avatar for user DJ Rogers
    Is there something wrong with the video? I watched it through, but I didn't get much, so I wanted to watch it again, but it keeps telling me I finished it and the next one auto plays.
    (0 votes)
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  • eggleston blue style avatar for user Pix
    I still don't understand how to factor the equation in the video even though I watched it 5 times. Especially when Mr. Khan factors the trinomial at in the video.
    (2 votes)
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    • blobby purple style avatar for user Bikram
      I will try to explain this to the best of my abilities so for (x^2-4x+4) the reason we can do (x-2)(x-2) is take for example (a+b)(a+b) that gives us a^2 + 2ab + b^2 now this pattern can be applied to the polynomial where x^2 is a^2 so a = x and 4 is b^2 so b = positive or negative 2 (remember a negative squared is always positive) and finally for the middle part 2ab that has to be -4x so let's see what to use (2)(x) now it's -4x so think what do I multiply 2x with to get -4x, that's -2 so 2ab = (2)(x)(-2) now we can plug in the value of a which is x and the value of b which is -2 (negative 2) into (a+b)(a+b) or (a+b)^2 which gives us (x-2)(x-2) or (x-2)^2, you can expand this and see you come to the same result by doing
      x(x)+x(-2)+(-2)(x)+(-2)(-2)=
      = x^2+(-2x)+(-2x)+4
      = x^2-4x+4
      , and this is how at the time you marked Mr.Khan did that, also if you want to get a better grasp try looking up the videos on "perfect square" and "difference of squares" where Mr.Khan goes into more debth!
      (1 vote)

Video transcript

- [Instructor] There are many videos on Khan Academy where we talk about factoring polynomials. What we're going to do in this video is do a few more examples of factoring higher degree polynomials. So let's start with a little bit of a warmup. Let's say that we wanted to factor six x squared plus nine x times x squared minus four x plus four. Pause this video and see if you can factor this into the product of even more expressions. All right, now let's do this together and the way that this might be a little bit different than what you've seen before is this is already partially factored. This polynomial, this higher degree polynomial, is already expressed as the product of two quadratic expressions but as you might be able to tell, we can factor this further. For example, six x squared plus nine x, both six x squared and nine x are divisible by three x. So let's factor out a three x here. So this is the same thing as three x times, three x times what is six x squared? Well, three times two is six and x times x is x squared and then three x times what is nine x? Well, three x times three is nine x and you can verify that if we were to distribute this three x, you would get six x squared plus nine x and then what about this second expression right over here? Can we factor this? Well, you might recognize this as a perfect square. Some of you might have said, hey, I need to come up with two numbers whose product is four and whose sum is negative four and you might say, hey, that's negative two and negative two and so this would be x minus two. We could write x minus two squared or we could write it as x minus two times x minus two. If what I just did is unfamiliar, I encourage you to go back and watch videos on factoring perfect square quadratics and things like that but there you have it. I think we have factored this as far as we can go. So now let's do a slightly trickier higher degree polynomial. So let's say we wanted to factor x to the third minus four x squared plus six x minus 24 and just like always, pause this video and see if you can have a go at it and I'll give you a little bit of a hint. You can factor in this case by grouping and in some ways it's a little bit easier than what we've done in the past. Historically, when we've learned factoring by grouping, we've looked at a quadratic and then we looked at the middle term, the x term of the quadratic and we broke it up so that we had four terms. Here, we already have four terms. See if you can have a go at that. All right, now let's do it together. So you can't always factor a third degree polynomial by grouping but sometimes you can so it's good to look for it. So when we see it written like this, we say okay, x to the third minus four x squared, is there a common factor here? Well, yeah, both x to the third and negative four x squared are divisible by x squared. So what happens if we factor out an x squared? So that's x squared times x minus four and what about these second two terms? Is there a common factor between six x and negative 24? Yeah, they're both divisible by six. So let's factor out a six here. So plus six times x minus four and now you are probably seeing the homestretch where you have something times x minus four and then something else times x minus four and so you can, sometimes I like to say undistribute the x minus four or factor out the x minus four and so this is going to be x minus four times x squared, x squared plus six and we are done.