Learn how to factor 5-term polynomials by grouping. Factor x^4-4x^3+2x^2-11x+12 By rational zero theorem: • How to factor a 5-term... By grouping: • How to factor a 5-term... The double-cross method: • How to factor a 5-term...
Factor x^4-4x^3+2x^2-11x+12 By rational zero theorem: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-yx2RetjV1Bo.html By grouping: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-55ufNfFofzY.html The double-cross method: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-B8dCd6PkHMY.html
"Obvious" is in the eye of the beholder. My first step on the first one was to factor out the x+1. For the way I think, that was the obvious starting point. Then I was stumped and I expected something clever. Nope. Just leave the x^3+x^2+1 unfactored without explanation. Somehow, I've never been good at knowing when to stop factoring a polynomial. For the second one, I made a mistake when I wrote it down, and broke it down into complex factors just for the satisfaction of factors with no power of x greater than 1. But I was wrong from the start, so the answer was nonsense. I might as well have shown what shade of orange the sky is.
For anybody looking for a great book on factoring, I recommend Schaum's College Algebra outline. It doesn't give problems exactly like this, but it gives you a thorough education in fairly advanced factoring methods. For the 2nd problem, I broke the 2x down to x + x, then I used factoring by grouping to get x^3(x + 1) + x(x + 1) + 1(x + 1), factoring out (x + 1), and my answer was (x + 1)(x^3 + x + 1). HIs method is probably a little more straightforward, but it's getting late and I somehow missed x^2 + 2x + 1, even though it was in the first problem.
Cant focus, instructor has too much sauce. Serious question: How do you go about factoring (x^3+x^2+1) if you need to find all the x intercepts (both real and imaginary)?
excellent... this video helped me a lot to solve a hard problem in V-T graphs.Although this factorization isn't tough..... i couldn't think of this at that moment until I saw your video.. thank you very much for this video.
I did it in a bit different way and came to the same answer x^4 + 2x^3 + x^2 + x + 1 = x^4 + x^3 + x^3 + x^2 + (x + 1) = x^3(x+1) + x^2(x+1) + (x+1) =(x^3+x^2 )(x+1) + (x+1) = (x+1)(x^3+x^2+1) It looks big as I have written almost every step.
what math do u learn this in. Im guessing thsi way isnt what im learnin yet in precal. I have to use potential rational roots way with synthetic division.
I don't why algebra books say 4-6 terms grouping to factor but never mention other cases I've been searching for anything about 5 terms factoring by grouping but only found this video.
@@arzaseb this is not an equation where X has an unknown value that could make x+1 be 0. It's just a polynomial, x has no value and x+1 can't be 0 it's just x+1.
Pretty much when you run out of ideas. For instance, in the last example, I know that the cubic must be factorable (all polynomials with degree > 2 definitely are), but I can't find a rational root or any easy angle of attack. Using a CAS to determine the roots of this, I get x=((sqrt(31)-3^(3/2))^(2/3)-2^(2/3))/(2^(1/3)*sqrt(3)*(sqrt(31)-3^(3/2))^(1/3)). I think it is safe to say we weren't meant to decompose the cubic further at this point...
I want a video on Indrajeet's Law which works for nth degree polynomial equation. we can find all real roots of nth degree polynomial equation easily by Indrajeet's Law, I want a video on Indrajeet's Law, if anyone wants make a video on Indrajeet's Law then first of all they should read this on Google.
f(x) = x⁴ + 3x + 20 has no real zeros. Minimum occurs at x = -∛¾ = -.908560296416..., where f(x) = 17.955739333064... Every even-degree polynomial with real coefficients can be factored into a product of quadratics with real coefficients, but finding those quadratics in general is not easy. Note that in general, their coefficients will not be rational, even when the original coefficients, like here, are integers. Fiddling around a bit, I find that, conveniently, this quartic _can_ be factored into a pair of quadratics with integer coefficients: x⁴ + 3x + 20 = (x² + 3x + 4)(x² - 3x + 5) I started by noting that because the cubic term is 0, the linear terms of the two quadratics must be ±a, for some number, a. Secondly, because the constant term is 20, the constant terms of the two quadratics must multiply to 20; so call them b and 20/b. Then you have: x⁴ + 3x + 20 = (x² + ax + b)(x² - ax + 20/b) = x⁴ + (b - a² + 20/b)x² + a(20/b - b)x + 20, . . . which means we get the two equations in a and b: 1. b - a² + 20/b = 0 ⇒ b² - a²b + 20 = 0 ⇒ b = ½(a² ± √[a⁴-80]) ⇒ 20/b = ½(a² ∓ √[a⁴-80]) 2. a(20/b - b) = 3 ⇒ ½a(∓2√[a⁴-80]) = ∓a√[a⁴-80] = 3 . . . A³ - 80A - 9 = 0, where A == a² This is where you can find that there's an integer solution . . A = 9 satisfies A³ - 80A - 9 = 0 : : : 729 - 80·9 - 9 = 0 So a = 3 . . . [taking the positive sign here, because reversing it merely swaps the two quadratic factors] and b = 4, 20/b = 5, and you get the factorization above. And sure enough, all four zeros are complex non-real. Fred
