# Math Help - Functions congruency proof help

1. ## 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.

2. Originally Posted by jconfer
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...