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A tricky remainder theorem problem

  1. Jun 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A polynomial P(x) is divided by (x-1), and gives a remainder of 1. When P(x) is divided by (x+1), it gives a remainder of 3. Find the remainder when P(x) is divided by (x^2 - 1)

    2. Relevant equations
    Remainder theorem

    3. The attempt at a solution
    I know that

    P(x) = (x-1)A(x) + 1
    and P(x) = (x+1)B(x) + 3

    But how would I relate to (x^2 -1)? I can multiply the two equations together to get (x^2 -1) but things get pretty messy.
     
  2. jcsd
  3. Jun 12, 2016 #2
    From your two equations we know P(1) = 1 and P(-1) = 3 . Since the divisor in question ([itex]x^2 - 1[/itex]) the remainder has degree < divisor i.e it is linear. Let it be Kx + L. The we get an equation [tex] P(x) = (x^2 - 1)f(x) + Kx + L [/tex] . Try to proceed from here.
     
  4. Jun 13, 2016 #3
    Why is the remainder is a linear expression? I can't catch your explanation.
     
  5. Jun 13, 2016 #4

    Ray Vickson

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    The remainder in ##a(x)/b(x)## is a polynomial ##r(x)## of degree strictly less than the degree of ##b(x)##. Basically, that is the _definition_ of "remainder".
     
  6. Jun 13, 2016 #5
    So that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.
     
  7. Jun 13, 2016 #6

    Ray Vickson

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    Go back and read post #2 again, then read your question in post #3.
     
  8. Jun 13, 2016 #7
    So it has something to do with the divisor right?
     
  9. Jun 13, 2016 #8

    rcgldr

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    To refresh the idea of the remainder theorem, take a look at

    http://en.wikipedia.org/wiki/Polynomial_remainder_theorem

    In this case you have P(1) = 1 (since P(x)/(x-1) has remainder 1), P(-1) = 3 (since P(x)/(x+1) has remainder 3). Can you think of a minimum degree P(x) that produces these results?
     
  10. Jun 13, 2016 #9
    A cubic polynomial?
     
  11. Jun 13, 2016 #10

    rcgldr

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    Ignore the fact that you will be looking for P(x) / (x^2-1). Can you think of a minimum degree polynomial P(x) such that P(-1) = 1 and P(1) = 3?
     
  12. Jun 13, 2016 #11
    Degree 3?
     
  13. Jun 13, 2016 #12

    rcgldr

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    Looking for a minimal degree for P(x). Start off with degree 1, is there a P(x) of degree 1 (ax + b) such that P(-1) = 1 and P(1) = 3? If not, try degree 2, and if not, try degree 3.
     
  14. Jun 13, 2016 #13
    Yup it appears that degree one works as well...
     
  15. Jun 13, 2016 #14

    rcgldr

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    So what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?
     
  16. Jun 13, 2016 #15
    ax+b, and I managed to solve it as (2-x).
     
  17. Jun 13, 2016 #16

    rcgldr

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    To follow up, this would mean that a general equation for P(x) = Q(x)(x-1)(x+1) - x + 2, where Q(x) can be any function of x, including Q(x) = 0.
     
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