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Find polynoms, with as least as power possible

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Find polynoms, with as least as power possible, so they will be competent for the equation.

    [tex](x^4+2x^3+x+1)A(x) + (x^4+x^3-2x^2+2x-1)B(x)=x^3-2x[/tex]

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2008 #2
    Please help.
     
  4. Mar 9, 2008 #3

    tiny-tim

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    Hi Theofilius! :smile:

    Just use trial and error …

    A and B scalars obviously doesn't work.

    So try A and B first-order, of the form A(x) = ax + b, B(x) = cx + d; if that doesn't work, try second-order, if that doesn't work …

    Good luck! :smile:
     
  5. Mar 9, 2008 #4
    Is there another way, much sure than this one ? :)
     
  6. Mar 9, 2008 #5

    HallsofIvy

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    There are two ways to do it- you can wave a magic wand or you can do the work! Why would you consider what Tiny Tim suggested "not sure"?

    A little thought might simplify the work: since there are no x4, x2 or constant terms on the right hand side, those will have to cancel out on the left.
     
  7. Mar 9, 2008 #6
    I mean, if it is [tex]x^100 + x^99[/tex], aren't any fixed way? If not its ok.
     
  8. Mar 9, 2008 #7

    tiny-tim

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    (erm … you have to use {} round the 100 or the 99, or it comes out like … well, like that!)

    Sorry, Theofilius, not understanding you … not even with the babel fish. :confused:
     
  9. Mar 9, 2008 #8
    Sorry, I meant if there any 100% way sure, is there any fixed way, so I can solve the equation withouth trying?
     
  10. Mar 9, 2008 #9

    tiny-tim

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    :rofl: Theofilius, nice try! :rofl:​

    … but there wouldn't be any point in your teachers setting these exercises if you could do them without trying, would there? :smile:
     
  11. Mar 9, 2008 #10
    Sorry, if I am misunderstud. I want to know is there any fixed principle for solving this equation? If not, its ok. In which form will be A(x). [tex]A(x)=ax^2+bx+c[/tex]. What about B(x)?
     
  12. Mar 9, 2008 #11

    tiny-tim

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    So far as I know, the "fixed principle" is the trial-and-error method I suggested.

    (It may also be possible to do it by dividing by [tex]x^3\,-\,2x[/tex] and then dealing with the remainders, but
    (a) I think that might take even longer for a relatively short problem like this, and
    (b) you're not good at dividing by polynomials, are you?)

    Try A(x) and B(x) both of the form ax + b first.

    If that doesn't work, try A(x) and B(x) both of the form [tex]ax^2+bx+c[/tex].

    If that doesn't work, … :smile:
     
  13. Mar 9, 2008 #12
    I tried with dividing, but it didn't worked. I will try with the other method.
     
  14. Mar 10, 2008 #13
    Ok, I tried to solve the system of equation, and it didn't work.
    [tex]A(x)=ax^2+bx+c[/tex] [tex]B(x)=ex^2+fx+g[/tex]
    Also I tried with
    [tex]A(x)=ax^2+bx+c[/tex] [tex]B(x)=ax^2+bx+c[/tex]
    and nothing.
     

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  15. Mar 10, 2008 #14
    There are 1000 of combinations, isnt any simpler way of doing this job?
     
  16. Mar 10, 2008 #15

    tiny-tim

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    You're almost there …

    Theofilius, you're giving up too easily …

    You're nearly half-way there, because your own equations immdiately eliminate some of the unknowns.

    You've written a + e = 0 and c - g = 0.

    So e = -a, and g = c.

    Now rewrite the other four equations, without e and g (for example the second line is b + a + f = 0) … :smile:
     
  17. Mar 10, 2008 #16
    it has no solution
     
  18. Mar 11, 2008 #17
    Really if some polynom is given, we should try (ax+b) for the 1st one, and then (cx+d) for the second one. If not, then (ax^2+bx+f), and for the 2nd one (cx+d) and vice versa, which is too complicated, and wasting time, there must be some other solution, less complicated than this one.
     
  19. Mar 12, 2008 #18
    I hope so that there is another way, because if it is given on some paper test, it will take me 30min. to solve it.
     
  20. Mar 12, 2008 #19
    By dividing the whole polynom with [itex]x^3-2x[/itex], we receive:

    [tex]((x+2) + \frac{2x^2+3x+1}{x^3-2x})A(x) + ((x+1) + \frac{4x-1}{x^3-2x})B(x)=1[/tex]

    Now, [tex]1=0*A(x)+\frac{1}{B(x)}B(x)[/tex]

    Now, we can make some system, what do you think?
     
    Last edited: Mar 13, 2008
  21. Mar 13, 2008 #20
    tiny-tim what do you think?
     
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