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Callisto Group Title

Tutor question #2 Factor: x^4 -7x^2 +1 (conditions: pen, paper, no calculator, done in 2 minutes) **

  • 2 years ago
  • 2 years ago

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  1. maheshmeghwal9 Group Title
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    May be we can take x^2 as "t" & then factor

    • 2 years ago
  2. maheshmeghwal9 Group Title
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    I mean\[(t^2-7t+1+5)-5\]

    • 2 years ago
  3. Callisto Group Title
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    If my brain still works, Then it becomes (t-6)(t-1)-5 and then ..?

    • 2 years ago
  4. ParthKohli Group Title
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    I should learn this :/

    • 2 years ago
  5. maheshmeghwal9 Group Title
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    ya it is not going to be factor like this:/ May be @satellite can help us:)

    • 2 years ago
  6. ParthKohli Group Title
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    It's @satellite73

    • 2 years ago
  7. ParthKohli Group Title
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    And he's offline

    • 2 years ago
  8. maheshmeghwal9 Group Title
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    oops sorry lol

    • 2 years ago
  9. maheshmeghwal9 Group Title
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    so we made only this much\[(x^2-6)(x^2-1)-5.\] now what should we do of {-5}?

    • 2 years ago
  10. Callisto Group Title
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    I don't think we should do it in this way...

    • 2 years ago
  11. maheshmeghwal9 Group Title
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    ya u r right @Callisto But I m thinking more on this.

    • 2 years ago
  12. bronzegoddess Group Title
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    you need to use synthetic division

    • 2 years ago
  13. Callisto Group Title
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    @bronzegoddess Can you further explain it?

    • 2 years ago
  14. maheshmeghwal9 Group Title
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    x^2-6 @ParthKohli

    • 2 years ago
  15. Callisto Group Title
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    @ParthKohli It's not like that.... and you made mistakes there too...

    • 2 years ago
  16. ParthKohli Group Title
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    I deleted it before you noticed :P

    • 2 years ago
  17. maheshmeghwal9 Group Title
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    lol

    • 2 years ago
  18. Callisto Group Title
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    I noticed before you deleted :P

    • 2 years ago
  19. maheshmeghwal9 Group Title
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    ya i too

    • 2 years ago
  20. ParthKohli Group Title
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    This degree 4 is eating my head >.<

    • 2 years ago
  21. maheshmeghwal9 Group Title
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    I m feeling giddy too

    • 2 years ago
  22. ParthKohli Group Title
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    @maheshmeghwal9 can you call Yogesh please?

    • 2 years ago
  23. maheshmeghwal9 Group Title
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    but he is sitting aside of me

    • 2 years ago
  24. maheshmeghwal9 Group Title
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    we 3 are all here:)

    • 2 years ago
  25. ParthKohli Group Title
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    Haha nice

    • 2 years ago
  26. bronzegoddess Group Title
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    okay, lets make a guess that (x-2) is a factor then we'll use synthetic division to actually se eif it one, its a trial and error method though: |dw:1339672516505:dw| so according to this division(x-2) isnt a factor because it gives us a remainder 11.

    • 2 years ago
  27. ParthKohli Group Title
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    But my wolfy says (x^2 - 3x + 1)(x^2 + 3x + 1)

    • 2 years ago
  28. Callisto Group Title
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    @bronzegoddess According to the word 'factor', I don't expect there is a remainder.

    • 2 years ago
  29. maheshmeghwal9 Group Title
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    ya @Callisto is right: D

    • 2 years ago
  30. Callisto Group Title
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    @ParthKohli I said no wolf. I got that answer from wolf too. But that doesn't help you solve it because you cannot use calculator. Just pen and paper

    • 2 years ago
  31. ParthKohli Group Title
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    Hmm but I was telling bronzegoddess that this can be factored. I wanna understand too

    • 2 years ago
  32. ParthKohli Group Title
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    LANA THE PROFESSOR :D

    • 2 years ago
  33. Callisto Group Title
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    Not using synthetic division like that - yes :) Lana~~~~~~~~~~~~

    • 2 years ago
  34. bronzegoddess Group Title
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    @Callisto as I said it is a trail and error, and i was giving you an example, of course you have to start with finding the possible factors using descartes rule of signs!

    • 2 years ago
  35. maheshmeghwal9 Group Title
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    We must want an exact & good method.

    • 2 years ago
  36. bronzegoddess Group Title
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    okay, let me do the whole thing again.

    • 2 years ago
  37. Callisto Group Title
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    @bronzegoddess If the factor is 2 terms like ax+b, I can do it. But in this case, it's not. So I don't know how. If you don't mind, can you teach me how to reach that answer?

    • 2 years ago
  38. maheshmeghwal9 Group Title
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    very good @ajprincess

    • 2 years ago
  39. maheshmeghwal9 Group Title
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    but how did she know about adding -3x^3 & something other. I want to hear from her.

    • 2 years ago
  40. maheshmeghwal9 Group Title
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    ya but if we didn't know the answer, then what to do It is the exact problem:/

    • 2 years ago
  41. ajprincess Group Title
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    ya will show that in a few minutes time.

    • 2 years ago
  42. maheshmeghwal9 Group Title
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    hmm..

    • 2 years ago
  43. bronzegoddess Group Title
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    first: number of possible roots is 4 because of the degree. second Descartes' Rule: number of +real roots = 1or zero, and -real roots= 1 or zero, imaginary=2 or 4 third: what are the possible roots; constant term/leading coefficient so \[\pm1/\pm1\] therefore your 4 possible roots are -1,+1 and 2 imaginary numbers. fourth: synthetic division to see if -1 and +1 are actual roots; |dw:1339673466435:dw| but since it gives us a remainder then it isn't a factor(x+1) then you do the same with(x-1) and if it also give you a remainder then all your factors are imaginary. Thats how I learnt it.

    • 2 years ago
  44. Callisto Group Title
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    Do you know when to use that method?

    • 2 years ago
  45. Callisto Group Title
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    @bronzegoddess

    • 2 years ago
  46. ajprincess Group Title
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    we use trial and error method when we need to factorise expressions with greater power

    • 2 years ago
  47. bronzegoddess Group Title
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    when looking for factors or zeros of polynomials

    • 2 years ago
  48. Callisto Group Title
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    @bronzegoddess please take a look at this: http://en.wikipedia.org/wiki/Factor_theorem We can apply that method when one of the factors is in the form (x-k) Now, in this case, no factors are in the form (x-k). So, it cannot be applied in this case.

    • 2 years ago
  49. Callisto Group Title
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    Of course you can keep using that method for checking, but you can't get a factor, I think.

    • 2 years ago
  50. bronzegoddess Group Title
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    @Callisto (x-k) aka (x-1)=0 is x=1 which you'll end up using as the divisor in synthetic division.

    • 2 years ago
  51. Callisto Group Title
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    Okay, to be simple, if you can get a factor using that method, please tell me.

    • 2 years ago
  52. bronzegoddess Group Title
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    my step by step explanation from before doesn't make sense?

    • 2 years ago
  53. Callisto Group Title
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    Sorry, to be honest, it makes no sense to me in this case. I truly understand that method and have been using to to solve more than 50 questions - at least. But in this case, I don't think it can help me get a factor. Sorry if I offend you. But please read the things I posted, and I hope you understand that.

    • 2 years ago
  54. Callisto Group Title
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    And thank you for your time explaining that to me.

    • 2 years ago
  55. ajprincess Group Title
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    I agree @Callisto u cant get the answer in ths method. I tried with this method at first but didnt get the answr.

    • 2 years ago
  56. Callisto Group Title
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    *have been using that method to solve. Typo. Sorry

    • 2 years ago
  57. bronzegoddess Group Title
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    Ah so you already know synthetic division, I am not offended at all. I just wished to help but unfortunately it's not working. Sorry for causing confusion then.

    • 2 years ago
  58. Callisto Group Title
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    @bronzegoddess Thanks for your time spent here and willingness to help! I appreciate your efforts put here.

    • 2 years ago
  59. bronzegoddess Group Title
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    But there is of 4 factors: option1: 2 real(a+ and a -) and 2imaginary or option2: they are all imaginary isn't it?

    • 2 years ago
  60. maheshmeghwal9 Group Title
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    May be of help @ some other time:)

    • 2 years ago
  61. Callisto Group Title
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    I just need real factors. The expression can be factorised, according to wolf

    • 2 years ago
  62. ajprincess Group Title
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    this sorts of questions can be solved without using calculators only by splitting the terms and by adding and removng additional terms. we will knw which terms to add and remove nly by workng many more questions of this type. I dnt thnk there is a specific way to solve this as far as I knw.

    • 2 years ago
  63. maheshmeghwal9 Group Title
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    Sorry about so long attachments

    • 2 years ago
  64. ParthKohli Group Title
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    I'm astonished. A question that looks as easy as simple factorization is confusing the polymaths.

    • 2 years ago
  65. maheshmeghwal9 Group Title
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    attachment is about factorizing:)

    • 2 years ago
  66. bronzegoddess Group Title
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    when you find an answer please do inform me :) thank you!

    • 2 years ago
  67. ajprincess Group Title
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    Bt seems so useful @maheshmeghwal9.

    • 2 years ago
  68. maheshmeghwal9 Group Title
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    thanx:) @ajprincess

    • 2 years ago
  69. ajprincess Group Title
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    yw.:)

    • 2 years ago
  70. maheshmeghwal9 Group Title
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    If u find answer @Callisto Plz inform me:)

    • 2 years ago
  71. experimentX Group Title
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    The general way to solve quartic equation is to Ferrari's method http://www.proofwiki.org/wiki/Ferrari%27s_Method ... for this particular method, assuming it has two quadratic factors, \[ (x^2 + ax + b)(x^2 + cx +d) = x^4 - 7x^2 + 1 \\ \text{ or, } \\ x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x + bd = x^4 - 7x^2 + 1 \\ \text{ Comparing coefficients } \\a+c = 0; \\b+d+ac=-7 \\ad+bc=0 \\ bd = 1 \] Now solving the values these equations we get the coefficients http://www.wolframalpha.com/input/?i=c%2Ba%3D0%2Cb%2Bd%2Bac%3D-7%2Cad%2Bbc%3D0%2C+bd%3D1 Hence we have our quadratic factors http://www.wolframalpha.com/input/?i=factorize+x^4+-+7x^2+%2B1 The other (numerical) method would be Newton's method http://en.wikipedia.org/wiki/Newton's_method The best rule would be to solve this type of equation simply is view analytically the function f(x) = x^4 -7x^2 + 1 eg, f(x) = x^4 +7x^2 +1 does not have any real solutions Besides Descartes's rule sign change and other techniques would be helpful if we were able to guess it's solution (using remainder theorem) then also we can easily factor out using synthetic division

    • 2 years ago
  72. bronzegoddess Group Title
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    @experimentX that was what i was saying all along :) is there a way to find the imaginary solutions then?

    • 2 years ago
  73. experimentX Group Title
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    yes there is, Ferarri's method would be the general method ... check that link it finds out both real and imaginary roots ... (it's a lot messy though ... even for simple problems) Easy method is to reduce it into quadratic factors or, reduce into cubic (x-a)(x^3+bx^2+cx+d) = 0 and use Cardano's method <--- this is messy too (x^2+bx+c)(x^2+dx+e) = 0 <--- this is the best you can expect from that equation as you know the roots are imaginary if 4ac > b^2 for quadratic factors.

    • 2 years ago
  74. maheshmeghwal9 Group Title
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    Hmm.... I heard of Ferrari today "A car also comes of help to maths{LOL}"

    • 2 years ago
  75. Callisto Group Title
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    Thanks

    • 2 years ago
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