2.2. Division of Polynomials 57
where q(t) is a polynomial and k a constant polynomial over D.(i) Show that degq = (degp) - 1.
(ii) By making the substitution t = c, verify that k = p(c).- Factor Theorem. Let c belong to an integral domain D and p(t) be
any polynomial over D. Show that (t - c) divides p(t) if and only if
p(c) = 0. - Let r be a zero of the polynomial p(t) over an integral domain, so
that for some polynomial q(t), p(t) = (t - r)q(t). Prove that, ifs # r,
then s is a zero of p(t) if and only ifs is a zero of q(t). - (a) Suppose rl, r2,... , rk are distinct zeros of a polynomial p(t) over
an integral domain. Show that there exists a polynomial q(t) for
which p(t) = (t - ?y)(t - r2)... (t - rk)q(i!).
(b) Prove that the number of distinct zeros of a nonzero polynomial
over an integral domain cannot exceed its degree.
(c) Verify that the polynomial t2-5t+6 over 212 has more than two
zeros. This example shows that (b) may fail when the condition
that the coefficients belong to an integral domain is dropped. - Let Q and b be two distinct zeros of a polynomial f(t), so that, for
some polynomials u(t) and v(t),
f(t) = (t - a)u(t) = (t - b)v(t).
Prove that the remaining zeros of f(t) are the solutions of the equa-
tion
u(t) - v(t) = 0.- Consider the equation -t4 - 51 - 6 = 0.
(a) By inspection, determine two integer solutions.
(b) Use Exercise 5 to determine two other solutions of the equation.- Exercise 1 treats the case when the divisor is of degree 1, in which case
we find that the remainder is a constant. Can we talk about division
by polynomials of degree exceeding l? Consider the possibility of
dividing the polynomial
f(t) = 4t5 - 3t4 - 7t2 + 6
by the polynomialg(t) = t3 + 7t2 + 3t - 2.