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Callisto

  • 2 years ago

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

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

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

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

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

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

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

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

  8. maheshmeghwal9
    • 2 years ago
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    oops sorry lol

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

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

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

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

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

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

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

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

  17. maheshmeghwal9
    • 2 years ago
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    lol

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

  19. maheshmeghwal9
    • 2 years ago
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    ya i too

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

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

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

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

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

  25. ParthKohli
    • 2 years ago
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    Haha nice

  26. bronzegoddess
    • 2 years ago
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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.

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

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

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

  30. Callisto
    • 2 years ago
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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

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

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

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

  34. bronzegoddess
    • 2 years ago
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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!

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

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

  37. Callisto
    • 2 years ago
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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?

  38. maheshmeghwal9
    • 2 years ago
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    very good @ajprincess

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

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

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

  42. maheshmeghwal9
    • 2 years ago
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    hmm..

  43. bronzegoddess
    • 2 years ago
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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.

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

  45. Callisto
    • 2 years ago
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    @bronzegoddess

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

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

  48. Callisto
    • 2 years ago
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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.

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

  50. bronzegoddess
    • 2 years ago
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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.

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

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

  53. Callisto
    • 2 years ago
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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.

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

  55. ajprincess
    • 2 years ago
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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.

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

  57. bronzegoddess
    • 2 years ago
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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.

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

  59. bronzegoddess
    • 2 years ago
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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?

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

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

  62. ajprincess
    • 2 years ago
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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.

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

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

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

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

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

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

  69. ajprincess
    • 2 years ago
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    yw.:)

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

  71. experimentX
    • 2 years ago
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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

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

  73. experimentX
    • 2 years ago
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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.

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

  75. Callisto
    • 2 years ago
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    Thanks

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