If $a/b$ is a root of a polynomial with integer coefficients, show that $a$ divides the constant term and $b$ divides the leading coefficient?

Question

Let $p(x) = c_0 + c_1x + \cdots + c_d x^d$ be a polynomial with integer coefficients $c_0, \ldots, c_d \in \mathbb{Z}$ and $c_d \neq 0$, so $p(x)$ has degree $d$. Let $a/b$ be a rational root of $p(x)$ written in lowest terms. How do I see that $a \mid c_0$ and $b \mid c_d$?

Answer 1

We know that $$c_0 + c_1\frac{a}{b}+ \cdots c_d(\frac{a}{b})^d = 0$$ is true. Multiplying both sides by $b^d$, we get $$c_0b^d + c_1 a b^{d-1}+ \cdots c_da^d = 0.$$ Right side is divisible by $a$ and $b$ so left side is also divisible by $a$ and $b$. I believe you can finish it from here!

Answer 2

Suppose $\frac ab$ is a root of $P(x)$ and $\gcd(a,b) = 1.$ i.e. $\frac ab$ is in lowest terms.

$P(\frac ab) = c_0 + c_1 \frac ab+c_2 \frac {a^2}{b^2} \cdots + c_d\frac {a^d}{b^d} = 0\\ c_1 \frac ab+c_2 \frac {a^2}{b^2} \cdots + c_d\frac {a^d}{b^d} = -c_0$

Every term on the left side of the equals sign is divisble by $a.$ Therefore $c_0$ is divisible by $a.$

Multiply trough by $b^d$ and rearrange the terms.

$c_d a^d = - c_0 b^d - c_1 a b^{d-1} \dots -c^{d-1} a^{d-1}b$

And similarly, every term on the right is divisible by $b,$ therefore $c_d$ is divisible by $b.$

Answer 3

First note you can assume $a$ and $b$ are coprime ($gcd(a,b)$=1). By substitution $x=\frac{a}{b}$ you get

$c_0+c_1\frac{a}{b}+\dots+c_d\frac{a^d}{b^d}=0$

Then multiply by $b^d$

$c_0b^d+c_1ab^{d-1}+\dots+c_da^d=0$

Now you see that $a$ divides the RHS, so it must divide LHS. But since it divides every summand except possibly $c_0b^d$, it must also divide this one. But $\gcd(a,b^n)=1$, so it must be $a\,|\,c_0$. By the same reasoning (swap $a$ and $b$, $c_0$ and $c_d$) you get $b\,|\,c_d$.

EDIT: I'm sorry for repeating the above posts, they weren't there when I started replying. I believed the website would have informed me before posting if other answer appeared while I was writing.

Answer 4

Here's a modular proof: $\,\bmod a\!:\ b^{-1}$ exists by $\,(a,b)=1,\,$ so $\,a/b\equiv ab^{-1}\equiv 0,\,$ so

$$\ {\rm mod}\ a\!:\,\ 0 \equiv p(a/b) \equiv p(0)\ \Rightarrow\, a\mid p(0) = c_0$$

The reversed poly $\ \bar p(x) = x^d p(1/x)\ $ has root $\, b/a,\,$ hence $\ b\mid \bar p(0) = c_d\,$ by above.

Remark $\ $ The is usually called the Rational Root Test. The above proof presumes that you know that modular arithmetic (congruences) can be extended to fractions with denominators coprime to the modulus (see the discussion here), a property which is greatly clarified when one studies ring theory (the universal properties of fraction rings or localizations).