5.4 POLYNOMIALS 171If a polynomial with integer coefficients can be fa ctored into polyno
mials with rational coefficients, it can also be fa ctored into primitive
polynomials with integer coefficients.
(A polynomial with integer coefficients is called primitive if its coefficients share no
factors. For example, 3x^2 + 9x + 7 is primitive while 1O.x^2 - 5x + 15 is not.) See
Problem 7.1 .30 for some hints on proving Gauss's Lemma.
Since P(x) factors into the product of (3x-2) and another polynomial with integer
coefficients, the coefficient of the leading term of P(x) must be a multiple of 3 and the
coefficient of the final term must be a multiple of 2.
In general, assume that a polynomial P(x) with integral coefficients has a rational
zero x = alb, where a and b are in lowest terms. By the Factor Theorem and Gauss's
Lemma,
P(x) = (bx-a)Q(x),
where Q(x) is a polynomial with integer coefficients. This immediately gives us the
Rational Root Theorem:
If a polynomial P(x) with integral coefficients has a rational zero x =
alb, where a and b are in lowest terms, then the leading coefficient of
P(x) is a multiple of b, and the constant term of P(x) is a multiple of a.
In practice, the Rational Root Theorem is used not just to find zeros but also to
prove that zeros are irrational.
Example 5.4.4 If .x^2 - 2 has any rational zeros alb (in lowest terms), we must have
bll and a 1 2. Therefore the only possible rational zeros are ±2. Since neither 2 nor - 2
are zeros, we can conclude that .x^2 - 2 has no rational zeros. This is another way to
prove that V2 is irrational!
We can generalize the above reasoning when applied to monic polynomials. It is
an interesting criterion for irrationality, and should be noted as a tool:
Any rational zero of a monic polynomial must be an integer. Con
versely, if a number is not an integer but is a zero of a monic poly
nomial, it must be irrational.^13
We shall conclude the section with a rather hard problem which uses the monic
polynomial tool above plus several other ideas.
Example 5.4.5 Prove that the sum
\hOOI 2 + 1 + VlO022 + 1 + ... + V 20002 + 1
is irrational.
Solution: Our strategy is two-pronged: first, to show that the sum in question is
not an integer, and second, to show that it is a zero of a monic polynomial.
(^13) This statement can also be proven directly, without using the rational roots theorem (Problem 5.4. 13). If you
are stumped, look at Example 7.1.^7 on page 226.