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Help polynomial

  1. Apr 1, 2004 #1
    if p(x ) = x^6 + x^4 + x^2 + 1

    show that the solutions of the equation p(x ) = 0 are among the solutions of the equation x^8 - 1 = 0

    hence factorse p(x ) fully over R
     
  2. jcsd
  3. Apr 1, 2004 #2

    matt grime

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    Can you factorize x^8 - 1?
     
  4. Apr 1, 2004 #3
    no, but you can factorise p(x ) = x^6 + x^4 + x^2 + 1.

    although i have not got time and resources to do it, im in a physics lesson.

    and why do you use p(x)= ? i always thaught it was supposed to be f(x)=. although it doesnt matter
     
  5. Apr 1, 2004 #4
  6. Apr 1, 2004 #5

    matt grime

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    well, you can factorize x^8-1, and it's because you can do that easily that you can factorize p(x) {p is just a label, it doesn' mean anything, Rattis, if you were taught always to use f how did you cope if there was more than one object to label?}. You don't need any (extra special) resources, you can do it in your head.
     
  7. Apr 1, 2004 #6
    yeah i think so x^8-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)

    but it seems useless



    and thx chen thats a good way solve it, but i was asked to solve hence by x^8-1
     
  8. Apr 1, 2004 #7
    You can factorize [tex]x^8 - 1[/tex] easily using [tex]a^2 - b^2 = (a + b)(a - b)[/tex] repeatedly.
     
  9. Apr 1, 2004 #8

    matt grime

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    But you can factorize it more than that.

    Hint

    let y=x^2, factorize y^4 - 1 as (y-1)(1+y+y^2+y^3), which you know. now put the x back in, what do you get? This is just saying that 1 and -1 are roots of x^8-1, and we factor them out to get...
     
  10. Apr 1, 2004 #9
    It's just:
    [tex]x^8 - 1 = (x^4 - 1)(x^4 + 1) = (x^2 - 1)(x^2 + 1)(x^4 + 1) = 0[/tex]
    And since p(x) can be factorized to:
    [tex]p_{(x)} = x^6 + x^4 + x^2 + 1 = (x^2 + 1)(x^4 + 1) = 0[/tex]
    .......
     
  11. Apr 1, 2004 #10
    wow surpised how came i never thinked abt that? ~~~
     
  12. Apr 2, 2004 #11
    resources as in time, and the fact that the teacher dosent notice me
     
  13. Apr 2, 2004 #12
    Actually, it is f(x), where f stands for function and the number inside your parenthesis (or x in this case) is the variable that the function is operating on.

    Paden Roder
     
  14. Apr 3, 2004 #13
    PRodQuanta, are you saying that it's wrong to use other letters than f to indicate a function...?
     
  15. Apr 3, 2004 #14

    matt grime

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    So, I've gto f(x) =sin(x), and I want another function to equal cos(x), and I must call that f as well must I?
     
  16. Apr 3, 2004 #15
    doesnt matter i guess, just incdicate group/function 01 =f(x) group/function 02 = g(x)
     
  17. Apr 3, 2004 #16
    Can you please define the derivative of the product of two different functions using f(x) alone?
     
  18. Apr 3, 2004 #17
    Hey, that's just what I have been taught. It must be wrong then. Sorry.

    Paden Roder
     
  19. Apr 3, 2004 #18
    Yeah, it's pretty wrong, considering there are functions like sin(x), det(A), log(x) and *gasp* [tex]\phi(n)[/tex]. ;)
     
  20. Apr 3, 2004 #19

    Hurkyl

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    If you don't see the sneaky way to factor it, the most straightforward approach is to just compute the 8 roots of x8-1 and then plug them all into 1 + x2 + x4 + x6.
     
  21. Apr 3, 2004 #20

    matt grime

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    Or sum a gp.
     
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