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Sum of the squares of remainders

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data

    The division of a polynomial f(x) by (x – 1)(x – 2) has remainder x + 1. If the remainder of the division of f(x) by (x – 1) and (x – 2) are, respectively, a and b. Then what is a^2 + b^2?

    2. Relevant equations

    I guess the remainder theorem could be useful here.

    3. The attempt at a solution

    f(x) = q(x)(x – 1) + a; where q(x) is the quotient of the division of f(x) by (x – 1).

    f(x) = p(x)(x – 2) + b; where p(x) is the quotient of the division of f(x) by (x – 2).

    q(x)(x – 1) + a = p(x)(x – 2) + b

    I don't know how to move forward using the fact the remainder of the division of f(x) by (x – 1)(x – 2).
     
  2. jcsd
  3. Nov 1, 2012 #2

    haruspex

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    What happens if you consider f(1)?
     
  4. Nov 1, 2012 #3
    f(1) = a and f(2) = b; right? Which means that a = -p(x) + b and b = q(x) + a.

    So a^2 + b^2 = (-p(x) + b)^2 + (q(x) + a)^2. I still don't see how to use the fact the remainder of the division of f(x) by (x – 1)(x – 2).

    EDIT: Wait; a + p(x) = b and b = q(x) + a. So a + p(x) = q(x) + a and p(x) = q(x)!! Is that right?
     
  5. Nov 1, 2012 #4
    You have established that [itex]f(1) = a[/itex] and [itex]f(2) = b[/itex].
    But we also know that
    [itex]f(x) = g(x)(x-1)(x-2) + (x+1)[/itex]
    for some quotient [itex]g(x)[/itex].
    Can you compute what [itex]f(1)[/itex] and [itex]f(2)[/itex] are?
     
  6. Nov 1, 2012 #5
    f(1) = (1 + 1) and f(2) = (1 + 2). So a = 2 and b = 3. And a^2 + b^2 = 13.

    How do you guys see problems so clearly? Now that you pointed out everything became obvious, but before that I could never have seen that. Is that practice? I feel like I can grasp concepts really well, but when it comes to problem solving I can hardly interpret the question at first or second sight. How can I improve that?
     
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