Page 232 - DMTH403_ABSTRACT_ALGEBRA

Unit 23: Division of Algorithm

Now, since the degree of the remainder -5x- 4 is less than deg .(x +x+1), we stop the process. We Notes
2
get
x + x + 5x – x = (x + x + 1) (x + 4) – (5x + 4)
2
2
2
3
4
Here the quotient is x + 4 and the remainder is – (5x+4).
2
Now, let us see what happens when the remainder in the expression f = qg + r is zero.
Self Assessment
1. Let F be a field. Let f(x) and g(x) be two polynomials is f[x], with g(x)  0, then the
polynomial q(x) and r(x) an ...................
(a) unique (b) deficient

(c) finite (d) infinite
2. If deg f(x) < deg g(x) we can chosen q(x) = 0. Then f(x) = 0.g(x) + f(x) where degf(x) ...................
deg g(x).

(a) < (b) >
(c)  (d) 
3. x + x + 5x – x is equal to ...................
2
3
4
(a) (x + x + 1) (q(x) + r(x) is Q[x])
2
(b) (x + x + 1) (q (x) + r (x) in Q[x])
-1
-1
2
(c) (x + x + 1) (q(x) + r(x) in Q[x])
-1
2
(d) q(x) + q(x) + (x + x + 1) in Q[x]
-1
2
2
4. ................... theorem said that let F be a field, if F[x]  P[x] and b  F, then there exists a
unique polynomial q(x)  F[x] such that f(x) = (i - b) q(x) + F(b)
(a) remainder theorem (b) division algorithm
(c) contradiction theorem (d) division matrix
23.2 Summary

The division algorithm in F[x], where F is a field, which states that if f(x), g(x)  F(x),

g(x)  0, then there exist unique q(x), r(x)  F[x] with f(x) = q(x) g(x)+r(x) and deg r(x)
< deg g(x).
a F is a root of f(x)  F[x] iff (x–a) | f(x).
A non-zero polynomial of degree n over a field F can have at the most n roots.

23.3 Keywords

Division Algorithm: Let F be a field. Let f(x) and g(x) be two polynomials in F[x], with g(x)  0.

Remainder Theorem: Let F be a field. If f(x)  P[x] and b  F, then there exists a unique polynomial
q(x)  F[x] such that f(x) = (i-b) q(x)+f(b).

LOVELY PROFESSIONAL UNIVERSITY 225