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High order thinking

  1. Aug 3, 2012 #1
    this is a HOT question from polynomials .

    1)find the remainder when x^100 is divided by x^2 - 3x + 2

    another from real numbers

    2) The number of + divisors which divide 10^999 but not 10^998 are ______


    i tried them for sometime . but could not solve them .
     
    Last edited: Aug 3, 2012
  2. jcsd
  3. Aug 3, 2012 #2

    Curious3141

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    What have you tried?
     
  4. Aug 3, 2012 #3
    well , i dint really know where to begin , so i just tried some factorization of the first equation and then dividing x^100 by them , but that dint work out . I also tried out a load of thinking but dint get anywhere :( . ( as to the second , im clueless )
     
  5. Aug 3, 2012 #4

    Curious3141

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    Let's tackle the first, well, first.

    One way to do this sort of problem is to use polynomial long division. While that may be practical for a lower order dividend (numerator), the dividend here is [itex]x^{100}[/itex], which makes this infeasible. So we have to use another approach.

    The divisor here is a quadratic expression. Do you know what form the remainder will take? Would it be a constant (a number) like when you divide a polynomial by a linear divisor? Or something else?

    Start by factorising the divisor ([itex]x^2 - 3x + 2[/itex]).
     
    Last edited: Aug 3, 2012
  6. Aug 3, 2012 #5
    the form of remainder is not mentioned , and i tried the factorization , doesnt work
     
  7. Aug 3, 2012 #6

    Curious3141

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    The remainder will be a polynomial expression. The order of the polynomial is at most one less than the order of the divisor. So if your divisor is a linear expression like x + 5, then you will have a constant remainder (like 3 or -5 or 0). If your divisor is a quadratic expression, your remainder may be a linear expression like x + 2 or 3x - 5, or it may be a constant like 2 or -5. All these are just examples to illustrate the point.

    So you should start by assuming the remainder is a linear expression, ax + b, where a and b are constants. If a comes out to be 0, then you know the remainder is simply a constant.

    When P(x) is divided by a quadratic expression [itex](x-\alpha)(x-\beta)[/itex], you can express the result by this equation:

    [itex]P(x) = (x-\alpha)(x-\beta)Q(x) + (ax + b)[/itex], where the remainder is [itex]ax + b[/itex].

    So your first step is to find out [itex]\alpha[/itex] and [itex]\beta[/itex], which is why I asked you to factorise the quadratic divisor. Surely, you know how to do that!
     
  8. Aug 3, 2012 #7
    oh yes , i know that , i told you i tried that , but the equation cannot be solved :/
     
  9. Aug 3, 2012 #8

    Curious3141

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    So WHAT is the factorisation of the divisor???
     
  10. Aug 3, 2012 #9

    Curious3141

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    You say you know to factorise the quadratic divisor, and this is all I want to see. Once you do that, I can walk you through the rest of the solution. But you have to factorise the quadratic, which is a very basic step, which you say you know how to do.

    So: just factorise [itex]x^2 - 3x + 2[/itex].

    It's late where I am, so I'm going to have to turn in now. I'll check in again in the morning my time (about 8 hours' time) and help you, provided no one else has guided you to a solution by then.
     
  11. Aug 3, 2012 #10

    eumyang

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    OP, start by writing the two numbers as products of a power of 2 and a power of 5. In other words,
    [tex]10^{999} = 2^{(something)} \cdot 5^{(something)}[/tex]
     
  12. Aug 3, 2012 #11

    LCKurtz

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    If you can't factor that, why would you be expect to be able to solve this problem? Perhaps you should take a look in a high school algebra book and do a review.
     
  13. Aug 3, 2012 #12

    Curious3141

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    Yes. Either he knows how to factor a quadratic, in which case he should show his work, or he doesn't, in which case this problem is not suitable for him at this stage.
     
  14. Aug 4, 2012 #13
    guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/
     
  15. Aug 4, 2012 #14
    u mean 2^999 * 3^999 ... ( why did you write 'something' ? )
     
  16. Aug 4, 2012 #15

    eumyang

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    Because I wanted you to fill in (something) with a number. And by the way, the 2nd base should be 5, not 3. So,
    [itex]10^{999} = 2^{999} \cdot 5^{999}[/itex] and
    [itex]10^{998} = 2^{998} \cdot 5^{998}[/itex]

    You'll notice that 2999 is not a factor of 10998. In fact, any factor of 10999 that is divisible by 2999 cannot be a factor of 10998. So how many of these factors of 10999 would that make altogether? (And no, the answer to this question is not the final answer of the problem.)
     
  17. Aug 4, 2012 #16

    ehild

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    First find the remainder when you divide x100 by (x-1).


    (hint: is xn-1 divisible by (x-1)?)

    ehild
     
    Last edited: Aug 4, 2012
  18. Aug 4, 2012 #17

    gabbagabbahey

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    Take another look at Curious3141's post (#6). What do you get when you look at [itex]x=\alpha[/itex] and [itex]x=\beta[/itex]? Apply that to your polynomial.
     
  19. Aug 4, 2012 #18

    Curious3141

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    OK, finally, you've factored it (and shown it). (BTW, that's not an "equation", merely an expression).

    You must have tried the wrong approach, because it's very easy from here on.

    Use this equation I previously quoted:

    [itex]P(x) = (x-\alpha)(x-\beta)Q(x) + (ax + b)[/itex], where the remainder is [itex]ax + b[/itex].

    for the current problem as:

    [itex]x^{100} = (x-1)(x-2)Q(x) + (ax + b)[/itex]

    and successively put [itex]x = 1[/itex] and [itex]x = 2[/itex] to get two simple, linear simultaneous equations in [itex]a[/itex] and [itex]b[/itex]. Solve those, and just write down the remainder as [itex]ax + b[/itex], using what you've found.
     
    Last edited: Aug 4, 2012
  20. Aug 4, 2012 #19
    case 1 . 1^100 = (1)a + b => 1 = a+b
    case 2 . 2^100 = (2)a = b => 2^100 = 2a+b

    if i solve these for a and b , which value of x will i take for the answer ( 1 or 2 ) , as the remainder is ax+b.

    this is the part where i failed.
     
  21. Aug 4, 2012 #20
    2^999 , 5^999 ?
     
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