Log in Angel C. 5 years agoPosted 5 years ago. Direct link to Angel C.'s post “What are some _common_ re...” What are some common real world applications for this? • (26 votes) J 5 years agoPosted 5 years ago. Direct link to J's post “You may need to use facto...” 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! (76 votes) jerimiah.moore 4 years agoPosted 4 years ago. Direct link to jerimiah.moore's post “So I have watched video a...” So I have watched video after video and I also took notes. I even took some hints I still cannot get this please help. • (17 votes) Rachael 4 years agoPosted 4 years ago. Direct link to Rachael's post “The first example is set ...” The first example is set up like this: The second example is a little different: (30 votes) Draven Brown 5 years agoPosted 5 years ago. Direct link to Draven Brown's post “I'm still confused on thi...” I'm still confused on this still after watching an hour. • (24 votes) Veerrohit 4 years agoPosted 4 years ago. Direct link to Veerrohit's post “group the factors into tw...” group the factors into two groups of 2 and then find the common factor and factor it out. Then, you will have a common factor and you will get a binomial equation. (0 votes) LarissaAnne 5 years agoPosted 5 years ago. Direct link to LarissaAnne's post “At 1:57 he mentions there...” At • (7 votes) webuyanycar.com 2 years agoPosted 2 years ago. Direct link to webuyanycar.com's post “@LarissaAnne, they are ht...” @LarissaAnne, they are https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:common-factor/v/algebraic-factoring-by-greatest-common-monomial-factor, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:common-factor/v/factoring-and-the-distributive-property-3, and https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:common-factor/v/factoring-polynomial-with-area-model I hope these help! (3 votes) Arwa a year agoPosted a year ago. Direct link to Arwa's post “is there another way othe...” is there another way other than "Grouping" to solve the higher-degree polynomials? • (1 vote) him 10 months agoPosted 10 months ago. Direct link to him's post “Yes, there are several me...” Yes, there are several methods to solve higher-degree polynomials (polynomials of degree three or higher) other than grouping. The most common methods include: 1. *Factoring*: This method involves factoring the polynomial into simpler expressions that can be set to zero to find the roots (solutions). Factoring works well for polynomials with rational roots or when they can be factored into binomial or trinomial expressions. However, it becomes more challenging for higher-degree polynomials with no simple factorization. 2. *Root-Finding Algorithms*: For polynomials that cannot be easily factored, numerical root-finding algorithms like the Newton-Raphson method, bisection method, or the secant method can be used to approximate the roots. These methods iteratively refine an initial guess until an accurate root is found. 3. *Completing the Square*: This method is commonly used for solving quadratic equations (polynomials of degree 2) but can be extended to higher-degree polynomials in some cases. By completing the square, you can transform certain higher-degree polynomials into a quadratic form that can then be solved using the quadratic formula. 4. *Descartes' Rule of Signs*: This rule provides information about the possible number of positive and negative real roots of a polynomial by counting the sign changes in its coefficients. It doesn't directly give the roots but narrows down the possibilities. 5. *Graphical Methods*: For a rough estimation of the roots, you can plot the graph of the polynomial and identify the approximate values where it crosses the x-axis, representing the roots. 6. *Numerical Methods*: Besides root-finding algorithms, there are other numerical methods like polynomial interpolation and numerical integration techniques that can help approximate the roots of higher-degree polynomials. It's essential to note that there is no general algebraic formula (analogous to the quadratic formula for degree 2 polynomials) for finding roots of polynomials with degrees higher than 4. For these higher degrees, numerical methods are often used to approximate the roots. Additionally, for some higher-degree polynomials, finding exact solutions may not be feasible or practical, and numerical methods are the best approach. (13 votes) caldwelljt a year agoPosted a year ago. Direct link to caldwelljt's post “I stopped the video and t...” I stopped the video and took a crack at it and got your answer, but then I also decided just for, um, fun? to recombine them to get 6x^4-15x^3-12x^2+36x, then I saw i could factor back out 3x, so 3x(2x^3-5x^2-6x+12)... which I noticed 3x was both in the original answer 3x(2x+3)(x-2)^2... so I presume that the 2x^3-5x^2-6x+12 could (or would) factor back down into (2x+3)(x-2)^2.... but i'm sort of lost on this circle... is only it only considered LCM in one of those, both of those? (neither?)... i guess i just noticed we weren't technically looking for LCM, so maybe this was just an exercise in options? I feel like guy who's mastered the art of stapling only to staple all four corners of a stack of papers. =P • (5 votes) Alaisia a year agoPosted a year ago. Direct link to Alaisia's post “Is anyone else confused a...” Is anyone else confused about this concept?! As a hormonal teenager this is making me feel a little on edge and kind of frustrated. Explanation please!!😶 • (5 votes) soccergurl4499 a year agoPosted a year ago. Direct link to soccergurl4499's post “saamee” saamee (1 vote) Annie Ice 3 years agoPosted 3 years ago. Direct link to Annie Ice's post “I just literally don't un...” I just literally don't understand how he got the (x-2) (x-2) part? I just am watching all the videos I can and still don't understand polynomials... pls, help... • (3 votes) Andrzej Olsen 3 years agoPosted 3 years ago. Direct link to Andrzej Olsen's post “There are a couple things...” There are a couple things that are good to know for this. One is this little trick, used in the video: (x+a)^2 = x^2 + 2a + a^2 (x-2)*(x-2) is (x-2)^2, so you can do that like this: (x-2)^2 = (x + -2)^2 = x^2 + 2(-2) + (-2)^2 = x^2 - 4x + 4 The hard part is figuring out when that's possible to do backwards. For things that work all the time, you need to look at quadratics, which tackle this general form: ax^2 + bx + c Here's the link to where Khan Academy teaches all about that: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring Good luck! (3 votes) Ŧг๏รՇ ๏ฬɭ 3 years agoPosted 3 years ago. Direct link to Ŧг๏รՇ ๏ฬɭ's post “im confused on how he doe...” im confused on how he does it is there more then one way to factor this same problem? • (3 votes) Simon 2 years agoPosted 2 years ago. Direct link to Simon's post “Well, it depends. If ther...” 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. (2 votes) Peter Lai 4 years agoPosted 4 years ago. Direct link to Peter Lai's post “How do you factor express...” How do you factor expressions with a difference of squares pattern? Please elaborate as it only shows this type of problems in practice. • (3 votes) Ender Ramsby a month agoPosted a month ago. Direct link to Ender Ramsby's post “You factor difference of ...” You factor difference of squares pattern by first recognizing them. The equation must be in either the form x^2 + y^2 or a(x^2 + y^2). Then, once you have figured out a, x, and y, you can factor the expression into a(x - y)(x + y). Hope this helps! (1 vote)Want to join the conversation?
(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.
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).
1:57
Video transcript
- [Instructor] There aremany 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 alittle bit of a warmup. Let's say that we wantedto factor six x squared plus nine x times x squaredminus four x plus four. Pause this video and seeif you can factor this into the product of even more expressions. All right, now let's do this together and the way that this mightbe a little bit different than what you've seen before is this is already partially factored. This polynomial, thishigher 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 ninex are divisible by three x. So let's factor out a three x here. So this is the samething as three x times, three x times what is six x squared? Well, three times two is sixand x times x is x squared and then three x times what is nine x? Well, three x times threeis nine x and you can verify that if we were todistribute this three x, you would get six x squared plus nine x and then what about this secondexpression right over here? Can we factor this? Well, you might recognizethis as a perfect square. Some of you might have said, hey, I need to come up with two numbers whose product is four andwhose sum is negative four and you might say, hey, that'snegative 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 xminus two times x minus two. If what I just did is unfamiliar, I encourage you to goback and watch videos on factoring perfect squarequadratics and things like that but there you have it. I think we have factoredthis as far as we can go. So now let's do a slightly trickier higher degree polynomial. So let's say we wantedto 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'velearned factoring by grouping, we've looked at a quadratic and then we looked at the middle term, the x term of the quadraticand 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 factora 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 thethird minus four x squared, is there a common factor here? Well, yeah, both x to the thirdand negative four x squared are divisible by x squared. So what happens if wefactor out an x squared? So that's x squared times x minus four and what about these second two terms? Is there a common factorbetween 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 probablyseeing the homestretch where you have somethingtimes x minus four and then something else timesx minus four and so you can, sometimes I like to sayundistribute the x minus four or factor out the x minus four and so this is going to be xminus four times x squared, x squared plus six and we are done.
