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Prove. Let p/q be a rational fraction in lowest terms and

a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + ... + a_{n-1}x + a_{n} = 0

be a polynomial equation with integral coefficients. If p/q is a root of the equation, then p is a
factor of a_{n} and q is a factor of a_{0}.

Proof. Let the given equation be

a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + ... + a_{n-2}x^{2} + a_{n-1}x + a_{n} = 0

If p/q is a root, then

a_{0} (p/q)^{n} + a_{1} (p/q)^{n-1} + a_{2} (p/q)^{n-2} + ... + a_{n-2} (p/q)^{2} + a_{n-1} (p/q) + a_{n} = 0

Multiplying both members by q^{n}, we obtain

1) a_{0}p^{n} + a_{1}p^{n-1}q + a_{2}p^{n-2}q^{2} + ... + a_{n-2}p^{2}q^{n-2} + a_{n-1}pq^{n-1 }+ a_{n}q^{n} = 0

Let us now write 1) as

a_{0}p^{n} + a_{1}p^{n-1}q + a_{2}p^{n-2}q^{2} + ... + a_{n-2}p^{2}q^{n-2} + a_{n-1}pq^{n-1 }= -a_{n}q^{n}

Written in this form, it is clear that p, being a factor of every term of the left member of the
equation, must divide a_{n}q^{n}. Because p/q is expressed in lowest terms, no factor of p will divide q.
Consequently, p must divide a_{n}. Thus the first part of our assertion is proved.

Let us now write 1) as

a_{1}p^{n-1}q + a_{2}p^{n-2}q^{2} + ... + a_{n-2}p^{2}q^{n-2} + a_{n-1}pq^{n-1 }+ a_{n}q^{n} = -a_{0}p^{n}

Using the same type reasoning as above, it follows that q must divide a_{0}.

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