Definition 1

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

f

(*x*) = *anx^n*+ *an-1**x^n-*1 + *... *+ *a*2*x^*2 + *a*1*x *+ *a*0*;*

where

an; an-

1*;...; a*2*; a*1*; a*0 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 to*g*(*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 + 3*x^*2 -2*x *+ 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.