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On polynomials

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A cubic polynomial gives remainders (5x + 4) and (12x -1) when divided by x^2 - x + 2 and x^2 + x - 1 respectively. Find the polynomial.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 12, 2008 #2

    Hurkyl

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    I presume that you seek help with this problem, rather than merely intending to duplicate your textbook?

    Well, we can't help you effectively without knowing what you have tried and where you are stuck....
     
  4. Mar 12, 2008 #3
    Sorry, this is the first time I posted here. I'm lost in this problem. I tried starting solving it by using:

    P(x)/(x^2 - x + 2) = Q(x) + (5x + 4) and

    P(x)/(x^2 + x - 1) = R(x) + (12x - 1)

    but to no luck can't solve it... any suggestions?
     
  5. Mar 12, 2008 #4

    Hurkyl

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    Well, first, you haven't divided quite correctly. (e.g. 4/3 = 1 + 1/3, not 1 + 1)

    You also know some information about Q and R, don't you?

    Finally, do you see any way to combine those two equations to get more information?
     
  6. Mar 12, 2008 #5
    just the basic stuff i know, but when all the x's come in, they confuse me...

    i tried to make P(x) = ax^3 + bx^2 + cx + d and divide it with x^2 - x + 2 giving a remainder x(b + c) -a(x + 2) - 2b + d which is equal to (5x + 4), and then divided P(x) with the x^2 + x - 1 giving a remainder x(2a + c) - bx + d - a which is also equal to (12x - 1), then i tried to solve simultaneously, but there are just too many unknowns... don't know what to do next...
     
  7. Mar 12, 2008 #6

    HallsofIvy

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    No, that's wrong. P(x)/(x^2- x+ 2)= Q(x)+ (5x+4)/(x^2- x+ 2) and
    p(x)/(x^2+ x- 1)= R(x)+ (12x- 1)/(x^2+ x- 1).

    It might help to write it as P(x)= Q(x)(x^2- x+ 2)+ 5x+4 and P(x)= R(x)(x^2+ x-1)+ 12x-1.
    You also know that P is cubic so Q and R must be linear. P(0)= 2Q(0)+ 4= -R(0)- 1 and P(1)= 2Q(1)+ 9= R(x)+ 11. Two points should be enough to determin a linear equation.
     
  8. Mar 12, 2008 #7

    Hurkyl

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    I see four dimensions of equations and four unknowns; there shouldn't be a problem. Don't forget that, for two polynomials to be equal, their coefficients must be equal!
     
  9. Mar 13, 2008 #8

    This method looks easy but I don't get it. Please explain further.


    I tried again to solve it simultaneously resulting in 2cx + ax - 3a - 2b + 2d = 17x + 3, still far from the solution.
     
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