Chapter 4: Q. 98 (page 210)

Prove the Rational Zeros Theorem.
[Hint: Let $\frac{p}{q}$ , where p and q have no common factors except $1$ and $- 1$ , be a zero of the polynomial function $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ whose coefficients are all integers. Show that
$a_{n} p^{n} + a_{n - 1} p^{n - 1} q + \dots + a_{1} p q^{n - 1} + a_{0} q^{n}$
Now, because p is a factor of the first n terms of this equation, p must also be a factor of the term $a_{0} q^{n}$ . Since p is not a factor of q (why?), p must be a factor of $a_{0}$ . Similarly, q must be a factor $a_{n}$ .]

Short Answer

Expert verified

For a function $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ where each coefficient is integers and $\frac{p}{q}$ is the rational zero then

p is a factor of the constant term $a_{0}$

q is factor of leading coefficient $a_{n}$

Step by step solution

Step 1. Given data

The given function is

$f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$

zero of the function is $- 1$

the common factor of factors of leading coefficient and the constant term is one

Step 2. Zero of function

As zero of function is $- 1$

$\frac{p}{q} = - 1 p = - q p + q = 0$

Step 3. Proof

Substitute $p + q$ for x in the function

$f (p + q) = a_{n} p^{n} + a_{n - 1} p^{n - 1} q + \dots + a_{1} p q^{n - 1} + a_{0} q^{n}$

p is a factor of n terms so p will be a factor of $a_{0} q^{n}$

p and q have a common factor of $1$ only so p is a factor of $a_{0}$ only

Similarly,q is a factor of $a_{n}$

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Short Answer

Step by step solution

Step 1. Given data

Step 2. Zero of function

Step 3. Proof

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