math111_logo Appendix: Polynomial

3. Division of polynomials

Let a and b be integers, b ≠ 0. Then we can always find unique integers q, r, such that

a = qb + r, 0 ≤ r < |b|.

The number q is the quotient of a divided by b, and r is the remainder of the division. In case r = 0, we say b divides a, or b is a divisor of a, or a is divisible by b.

We have similar division for polynomials.

Let a(t) and b(t) be polynomials, b(t) ≠ 0. Then there are unique polynomials q(t), r(t), such that

a(t) = q(t)b(t) + r(t), 0 ≤ deg r < deg b.

Example Let a(t) = t4 + t3 + 3t2 + t + 1, b(t) = t2 - t + 1. We may carry out the following divisions.

  t2 + 2t + 4 quotient
t2 - t + 1 | t4 + t3 + 3t2 + t + 1
  t4 - t3 + t2
  2t3 + 2t2 + t
  2t3 - 2t2 + 2t
  4t2 - t + 1
  4t2 - 4t + 4
  remainder 3t - 3

Each time, by trying to match the highest degree term, the degree of the remaining term is decreased by at least one. The algorithm will stop after the degree of remaining term is < deg b. The result of the division is

t4 + t3 + 3t2 + t + 1 = (t2 + 2t + 4)(t2 - t + 1) + (3t - 3).

If the remainder r(t) = 0, then we have a(t) = q(t)b(t) for a polynomial q(t). Similar to the case of numbers, we say b(t) divides a(t), or b(t) is a divisor of a(t), or a(t) is divisible by b(t). Note that the equality a(t) = q(t)b(t) has the following implication

r is a root of b(t) ⇒ b(r) = 0 ⇒ a(r) = q(r)b(r) = 0.

This proves the following result.

b(t) divides a(t) ⇒ roots of b(t) are roots of a(t).

Dividing any polynomial p(t) by a degree 1 polynomial b(t) = t - r gives us

p(t) = q(t)(t - r) + c,

where c is a number. Thus p(r) = c, and

r is a root of p(t) ⇔ c = 0 ⇔ p(t) = q(t)(t - r).

We just proved the following result.

r is a root of p(t) ⇔ p(t) = (t - r)q(t) for another polynomial q.


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