Elementary Functions 261
The factorization (5.14-3) shows that a polynomial of degree n has ex-
actly n zeros or roots (counting multiplicities). In fact, the substitution ofany other value j3 i= CXj (j = 1, ... , n) gives
P(/3) = (/3 - a1)(/3 - a2) · · · (/3 - CXn)an -=/= 0
A polynomial is said to be of apparent degree n if the value of the lead-
ing coefficient an is not yet known. For example, the integral algebraic
expression
(a - b)z + a(b-z) + b(z -a)
reduces to a polynomial of apparent degree 1.
The following theorem holds:
Theorem 5.11
(5.14-4)If a polynomial P(z) of apparent degree n ;::::: 1 has more than n zeros,
then each coefficient ao, ai, ... , an is zero; i.e., the polynomial is identically
zero.
Proof Suppose that there are nonzero coefficients, and let ak(l ::=; k ::=; n)
be the first one. Then the polynomial is of degree k and equal to zero for
more than k values of z, which is impossible. Hence ai = · · · = an = 0,
so that P(z) = a 0 (a constant polynomial), and since it vanishes for some
values of z, we have a 0 = 0 also.
Example The expression (5.14-4) is identically zero, since it vanishes for
z = a and z = b. ·
Corollary 5.1 (Identity Principle for Polynomials) If two polynomials
P(z) = a 0 +a1z+ .. · +anzn and Q(z) = bo +b1z+ · · · + bnzn assume equal
values for more than n values of z, they are identical; i.e., the coefficients
of corresponding powers of z are equal: ak = bk (0 ::=; k :=:; n).
The identity principle is the basis for the so-called method of undeter-
mined coefficients in elementary algebra.
Next, we wish to show that if P(z) is a polynomial of degree n;::::: 1, then
limz-+oo P( z) = oo. In fact, from
we have