Page 231 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Thus, we have proved the uniqueness of q(x) and r(x) in the expression f(x) = q(x) g(x)+r(x). Here
q(x) is called the quotient and r(x) is called the remainder obtained on dividing f(x) by g(x).
Now, what happens if we take g(x) of Theorem 1 to be a linear polynomial? We get the remainder
theorem. Before proving it let us set up some notation.

Notation: Let R be a ring and f(x)  R[x]. Let
f(x) = a + a x + ... +a x .
n
0
l
n
Then, for any r  R, we define
that is, f(r) is the value of f(x) obtained by substituting r for x.
Thus, if f(x) = 1 + x + x  Z[x], then
2
f(2) = 1 + 2 + 4 = 7 and f(0) = 1 + 0 + 0 = 1.
Let us now prove the remainder theorem, which is a corollary to the division algorithm.
Theorem 2 (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).
Proof: Let g(x) = x-b. Then, applying the division algorithm to f(x) and g(x), we can find unique
q(x) and r(x) in F[x], such that
f(x) = q(x)g(x) + r(x)
= q(x) (x – b) + r(x), where deg r(x) < deg g(x) = 1.
Since deg r(x) < 1, r (x) is an element of F, say a.

So, f(x) = (x - b)q(x) + a,
Substituting b for x, we get
f(b) = (b – b) q(b) + a
= 0.q(b) + a= a

Thus, a = f(b).
Therefore, f(x) = (x-b) q(x)+f(b).
Note that deg f(x) = deg(x-b)+deg q(x) = l+deg q(x).
Therefore, deg q(x) = deg f(x)-1.
Let us apply the division algorithm in a few situations now.

Example: Express x + x + 5x – x as
4
2
3
(x + x + 1) q(x) + r(x) in Q[x].
2
Solution: We will apply long division of polynomials to solve this problem.

4
2
3
2
x + x + 1) x + x + 5x  x
3
4
x  x  x 2
2
4x  x
2

4x  4x 4

 5x 4
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