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Find the remainder of a division

  1. May 13, 2007 #1
    Hi all!

    Could anyone help telling me the way to find the remainder of the following divisions:

    1. (x^2006+x^1996+x^1981+x+1):(x^2-1)

    2. (x2+x3+x5+1) : [(x-1)(x-2)]

    Last edited: May 13, 2007
  2. jcsd
  3. May 13, 2007 #2


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    Saying that "P(x) divided by U(x) is equal to quotient Q(x) with remainder R(x) means that P(x)= U(x)*Q(x)+ R(x)." Of course, R(x) has lower degree than U(x) so that, for example, if U is linear, then R(x)= r, a number. In both of your examples, U(x) is quadratic so R(x) is linear. It is also true that, if U(a)= 0 then P(a)= R(a).

    So: for the first problem, Q(x)= x2- 1 which has zeroes 1 and -1. You know that the remainder is a linear function, R(x)= ax+ b, such that R(1)= 1^2006+1^1996+1^1981+1+1= ? and R(-1)= (-1)^2006+(-1)^1996+(-1)^1981+(-1)+1= ?. Two points are sufficient to determine a and b. Same idea for the second problem.
  4. May 13, 2007 #3

    There may be a mistake in your reasoning. U(x) equals x^2-1 which has two roots at 1 and -1, not Q(x).
  5. May 13, 2007 #4

    D H

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    There is a typo in his reasoning, not a mistake. Substitute Q(x) with U(x) in Hall's write-up. It doesn't matter what Q(x) is. It is just some polynomial. If U(x) is zero for some particular value of x, the product Q(x)*U(x) will also be zero at that value of x.

    For a polynomial divided by a quadratic, you know that the remainder R(x) must be of the form ax+b. Since P(x) = Q(x)*U(x) + R(x), P(x) and R(X) must be equal at the zeros of U(x). Evaluating P(x) at the the zeros of U(x) provides the information needed to deduce the form of R(x).
  6. May 13, 2007 #5
    Oh, now I've got to the point:
    P(x) - R(x)=U(x).Q(x)
    At x=1 and x=-1, the left-hand side equals zero ..

    Thank you HallsofIvy and DH.
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