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Callisto

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

  • one year ago
  • one year ago

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

    • one year ago
  2. maheshmeghwal9
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    I mean\[(t^2-7t+1+5)-5\]

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

    • one year ago
  4. ParthKohli
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    I should learn this :/

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

    • one year ago
  6. ParthKohli
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    It's @satellite73

    • one year ago
  7. ParthKohli
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    And he's offline

    • one year ago
  8. maheshmeghwal9
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    oops sorry lol

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

    • one year ago
  10. Callisto
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    I don't think we should do it in this way...

    • one year ago
  11. maheshmeghwal9
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    ya u r right @Callisto But I m thinking more on this.

    • one year ago
  12. bronzegoddess
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    you need to use synthetic division

    • one year ago
  13. Callisto
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    @bronzegoddess Can you further explain it?

    • one year ago
  14. maheshmeghwal9
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    x^2-6 @ParthKohli

    • one year ago
  15. Callisto
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    @ParthKohli It's not like that.... and you made mistakes there too...

    • one year ago
  16. ParthKohli
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    I deleted it before you noticed :P

    • one year ago
  17. maheshmeghwal9
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    lol

    • one year ago
  18. Callisto
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    I noticed before you deleted :P

    • one year ago
  19. maheshmeghwal9
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    ya i too

    • one year ago
  20. ParthKohli
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    This degree 4 is eating my head >.<

    • one year ago
  21. maheshmeghwal9
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    I m feeling giddy too

    • one year ago
  22. ParthKohli
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    @maheshmeghwal9 can you call Yogesh please?

    • one year ago
  23. maheshmeghwal9
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    but he is sitting aside of me

    • one year ago
  24. maheshmeghwal9
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    we 3 are all here:)

    • one year ago
  25. ParthKohli
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    Haha nice

    • one year ago
  26. bronzegoddess
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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.

    • one year ago
  27. ParthKohli
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    But my wolfy says (x^2 - 3x + 1)(x^2 + 3x + 1)

    • one year ago
  28. Callisto
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    @bronzegoddess According to the word 'factor', I don't expect there is a remainder.

    • one year ago
  29. maheshmeghwal9
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    ya @Callisto is right: D

    • one year ago
  30. Callisto
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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

    • one year ago
  31. ParthKohli
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    Hmm but I was telling bronzegoddess that this can be factored. I wanna understand too

    • one year ago
  32. ParthKohli
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    LANA THE PROFESSOR :D

    • one year ago
  33. Callisto
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    Not using synthetic division like that - yes :) Lana~~~~~~~~~~~~

    • one year ago
  34. bronzegoddess
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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!

    • one year ago
  35. maheshmeghwal9
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    We must want an exact & good method.

    • one year ago
  36. bronzegoddess
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    okay, let me do the whole thing again.

    • one year ago
  37. Callisto
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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?

    • one year ago
  38. maheshmeghwal9
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    very good @ajprincess

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

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

    • one year ago
  41. ajprincess
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    ya will show that in a few minutes time.

    • one year ago
  42. maheshmeghwal9
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    hmm..

    • one year ago
  43. bronzegoddess
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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.

    • one year ago
  44. Callisto
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    Do you know when to use that method?

    • one year ago
  45. Callisto
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    @bronzegoddess

    • one year ago
  46. ajprincess
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    we use trial and error method when we need to factorise expressions with greater power

    • one year ago
  47. bronzegoddess
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    when looking for factors or zeros of polynomials

    • one year ago
  48. Callisto
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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.

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

    • one year ago
  50. bronzegoddess
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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.

    • one year ago
  51. Callisto
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    Okay, to be simple, if you can get a factor using that method, please tell me.

    • one year ago
  52. bronzegoddess
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    my step by step explanation from before doesn't make sense?

    • one year ago
  53. Callisto
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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.

    • one year ago
  54. Callisto
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    And thank you for your time explaining that to me.

    • one year ago
  55. ajprincess
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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.

    • one year ago
  56. Callisto
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    *have been using that method to solve. Typo. Sorry

    • one year ago
  57. bronzegoddess
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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.

    • one year ago
  58. Callisto
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    @bronzegoddess Thanks for your time spent here and willingness to help! I appreciate your efforts put here.

    • one year ago
  59. bronzegoddess
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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?

    • one year ago
  60. maheshmeghwal9
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    May be of help @ some other time:)

    • one year ago
  61. Callisto
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    I just need real factors. The expression can be factorised, according to wolf

    • one year ago
  62. ajprincess
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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.

    • one year ago
  63. maheshmeghwal9
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    Sorry about so long attachments

    • one year ago
  64. ParthKohli
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    I'm astonished. A question that looks as easy as simple factorization is confusing the polymaths.

    • one year ago
  65. maheshmeghwal9
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    attachment is about factorizing:)

    • one year ago
  66. bronzegoddess
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    when you find an answer please do inform me :) thank you!

    • one year ago
  67. ajprincess
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    Bt seems so useful @maheshmeghwal9.

    • one year ago
  68. maheshmeghwal9
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    thanx:) @ajprincess

    • one year ago
  69. ajprincess
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    yw.:)

    • one year ago
  70. maheshmeghwal9
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    If u find answer @Callisto Plz inform me:)

    • one year ago
  71. experimentX
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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

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

    • one year ago
  73. experimentX
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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.

    • one year ago
  74. maheshmeghwal9
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    Hmm.... I heard of Ferrari today "A car also comes of help to maths{LOL}"

    • one year ago
  75. Callisto
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    Thanks

    • one year ago
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