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Math Help - Functions congruency proof help

  1. #1
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    Functions congruency proof help

    Definition 1
    A polynomial with integer coefficients is a function f : R --> R of the form

    f
    (x) = anx^n+ an-1x^n-1 + ... + a2x^2 + a1x + a0;

    where
    an; an-
    1;...; a2; a1; a0 in the integers.

    Definition 3
    We say that a polynomial k divides a polynomial g if there is a polynomial f such that g(x) = f(x)k(x) for all

    x
    in the real numbers
    . In such a case we write k(x)/g(x).

    Definition 4
    Consider the polynomial k(x) = x -1. We say that two polynomials f and g are equivalent modulo x-1

    if
    (x -1)/[f(x) -g(x)]. In such a case, we write f(x) is congruent tog(x) (mod (x -1)).

    Example 1
    The polynomial k(x) = x -1 divides the polynomial g(x) = x^2 -1 because there is a polynomial f(x) = x + 1 such that g(x) = f(x)k(x); that is, x^2 -1 = (x + 1)(x -1).

    Example 2
    The polynomial k(x) = x-1 does not divide the polynomial g(x) = x^2 -2. When we divide (by long polynomial division) g(x) by k(x), we obtain a remainder of -1, not 0.



    Question:

    Let
    f(x) = x^3 + 3x^2 -2x + 6 and g(x) = 4. Show that f(x) is congruent to g(x) (mod (x-1)).


    Sorry about all the weird spacing i can't find where to fix it at.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jconfer View Post
    Definition 1
    A polynomial with integer coefficients is a function f : R --> R of the form
    f
    (x) = anx^n+ an-1x^n-1 + ... + a2x^2 + a1x + a0;
    where
    an; an-
    1;...; a2; a1; a0 in the integers.
    Definition 3
    We say that a polynomial k divides a polynomial g if there is a polynomial f such that g(x) = f(x)k(x) for all
    x
    in the real numbers
    . In such a case we write k(x)/g(x).
    Definition 4
    Consider the polynomial k(x) = x -1. We say that two polynomials f and g are equivalent modulo x-1
    if
    (x -1)/[f(x) -g(x)]. In such a case, we write f(x) is congruent tog(x) (mod (x -1)).
    Example 1
    The polynomial k(x) = x -1 divides the polynomial g(x) = x^2 -1 because there is a polynomial f(x) = x + 1 such that g(x) = f(x)k(x); that is, x^2 -1 = (x + 1)(x -1).
    Example 2
    The polynomial k(x) = x-1 does not divide the polynomial g(x) = x^2 -2. When we divide (by long polynomial division) g(x) by k(x), we obtain a remainder of -1, not 0.



    Question:
    Let
    f(x) = x^3 + 3x^2 -2x + 6 and g(x) = 4. Show that f(x) is congruent to g(x) (mod (x-1)).


    Sorry about all the weird spacing i can't find where to fix it at.
    f(x) is congruent to g(x)~(\mbox{mod }(x - 1)) means, (x - 1) divides f(x) - g(x)

    I suppose you can continue knowing that? just do the polynomial long division...
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