A polynomial with integer coefficients is a function f : R --> R of the form
(x) = anx^n+ an-1x^n-1 + ... + a2x^2 + a1x + a0;
1;...; a2; a1; a0 in the integers.
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
in the real numbers
. In such a case we write k(x)/g(x).
Consider the polynomial k(x) = x -1. We say that two polynomials f and g are equivalent modulo x-1
(x -1)/[f(x) -g(x)]. In such a case, we write f(x) is congruent tog(x) (mod (x -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).
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.
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.