Finding zeros of rational functions worksheet

Suppose \(c\) is a zero of \(f\) and \(c=\dfrac{p}{q}\) in lowest terms. This means \(p\) and \(q\) have no common factors. Since \(f(c) = 0\), we have

\(a_{n}\left(\dfrac{p}{q}\right)^{n}+a_{n-1}\left(\dfrac{p}{q}\right)^{n-1}+\ldots+a_{1}\left(\dfrac{p}{q}\right)+a_{0}=0\).

Multiplying both sides of this equation by \(q^{n}\), we clear the denominators to get

\(a_{n} p^{n}+a_{n-1} p^{n-1} q+\ldots+a_{1} p q^{n-1}+a_{0} q^{n}=0\)

Rearranging this equation, we get

\(a_{n} p^{n}=-a_{n-1} p^{n-1} q-\ldots-a_{1} p q^{n-1}-a_{0} q^{n}\)

Now, the left hand side is an integer multiple of \(p\), and the right hand side is an integer multiple of \(q\).

(Can you see why?) This means \(a_{n} p^{n}\) is both a multiple of \(p\) and a multiple of \(q\). Since \(p\) and \(q\) have no common factors, \(a_{n}\) must be a multiple of \(q\). If we rearrange the equation

\(a_{n} p^{n}+a_{n-1} p^{n-1} q+\ldots+a_{1} p q^{n-1}+a_{0} q^{n}=0\)

as

\(a_{0} q^{n}=-a_{n} p^{n}-a_{n-1} p^{n-1} q-\ldots-a_{1} p q^{n-1}\)

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