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

a0xn + a1xn-1 + a2xn-2 + ... + an-1x + an = 0

be a polynomial equation with integral coefficients. If p/q is a root of the equation, then p is a factor of an and q is a factor of a0.

Proof. Let the given equation be

a0xn + a1xn-1 + a2xn-2 + ... + an-2x2 + an-1x + an = 0

If p/q is a root, then

a0 (p/q)n + a1 (p/q)n-1 + a2 (p/q)n-2 + ... + an-2 (p/q)2 + an-1 (p/q) + an = 0

Multiplying both members by qn, we obtain

1)        a0pn + a1pn-1q + a2pn-2q2 + ... + an-2p2qn-2 + an-1pqn-1 + anqn = 0

Let us now write 1) as

a0pn + a1pn-1q + a2pn-2q2 + ... + an-2p2qn-2 + an-1pqn-1 = -anqn

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

Let us now write 1) as

a1pn-1q + a2pn-2q2 + ... + an-2p2qn-2 + an-1pqn-1 + anqn = -a0pn

Using the same type reasoning as above, it follows that q must divide a0.

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