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Here is the problem I am currently working on:

Suppose $\frac{a}{b}$ is a rational zero of the equation $$ a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n = 0,$$

where $a$ and $b$ are relatively prime integers and $a_0,a_1,...,a_n$ are arbitrary integers. Show that $a$ is a divisor of $a_n$ and that $b$ is a divisor of $a_0$.

I have no idea how to even approach this proof. What I do know is that $a$ and $b$ are relatively prime integers. In other words, their greatest common divisor is $1$ (so here we have gcd$(a,b) = 1$).

EDIT:

Originally, the textbook begins the question as

Suppose that a/b is a rational zero of the equation...

I just put it here just to follow the format of the textbook. I believe I'm suppose to split the fraction $\frac{a}{b}$ as $$a * \frac{1}{b},$$

since the problem asks to show that $a$ is a divisor of $a_n$ and that $b$ is a divisor of $a_0$.

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Since $\frac{a}{b}$ is a solution to the equation, we have: $$a_0\left(\frac{a}{b}\right)^n+a_1\left(\frac{a}{b}\right)^{n-1}+...+a_{n-1}\left(\frac{a}{b}\right)+a_n=0\\a_0(a)^n+a_1(a)^{n-1}b+...+a_{n-1}(a)b^{n-1}+a_nb^n=0$$ The last step results from multiplying both sides of the equation by $b^n$, which must be nonzero. Now factor out an $a$ from everything except the $a_n$ term, and move $a_nb^n$ to the other side. Use the fact that $\gcd(a,b)=1$ to show that $a$ must divide $a_n$.

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  • $\begingroup$ Okay Dave this should be enough to get me started. Anatoly's answer is also similar to yours as well. $\endgroup$ – John Smith Feb 18 '17 at 23:00
  • $\begingroup$ I believe the equation is multiplied by $b^n$, not $b$. $\endgroup$ – John Smith Feb 19 '17 at 0:36
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    $\begingroup$ You're right, that was a typo. The equation was multiplied by $b^n $, I just forgot to write that. Will fix. $\endgroup$ – Dave Feb 19 '17 at 0:37
  • $\begingroup$ No problem Dave. $\endgroup$ – John Smith Feb 19 '17 at 0:38
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Substituting $x $ with $\frac {a}{b} $ in the equation and multiplying both sides to $b^n $ you have

$$ a_0a^n+a_1a^{n-1} b+...+a_{n-1}a b^{n-1}+a_n b^n= 0$$

In this equation, all terms in the LHS except the last one contain the factor $a $, so $a_n $ must be divisibile by $a $. Similarly, all terms except the first one contain the factor $b $, so $a_0$ must be divisibile by $b $.

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  • $\begingroup$ Ahhh now I see it. And when you multiplied both sides by $b$, we know it's non-zero correct? $\endgroup$ – John Smith Feb 18 '17 at 22:58
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    $\begingroup$ Yes, we know that it is non-zero. $\endgroup$ – Anatoly Feb 18 '17 at 23:02
  • $\begingroup$ Okay just double checking. This should be enough to get me started. $\endgroup$ – John Smith Feb 18 '17 at 23:02

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