Roots of Polynomials over Z with Rational Roots

# Roots of Polynomials over Z with Rational Roots

We will now begin to investigate when we can determine whether a polynomial of a certain type has roots or not. We begin by looking at polynomials with coefficients in $\mathbb{Z}$ that have rational roots.

Theorem 1: Let $f \in \mathbb{Z}[x]$ with $f(x) = a_0 + a_1x + ... + a_nx^n$. If $\displaystyle{\frac{r}{s} \in \mathbb{Q}}$ is a root of $f$ with $\gcd (r, s) = 1$ then $r | a_0$ and $s | a_n$. |

**Proof:**Let $\displaystyle{\frac{r}{s} \in \mathbb{Q}}$ be a root of $f$. Then:

\begin{align} \quad f \left ( \frac{r}{s} \right ) = a_0 + a_1 \frac{r}{s} + ... + a_n \frac{r^n}{s^n} = 0 \end{align}

- We multiply both sides by $s^n$ to get:

\begin{align} \quad a_0s^n + a_1rs^{n-1} + ... + a_{n-1}r^{n-1}s + a_nr^n = 0 \end{align}

- From above since $s | a_0s^n$, $s | a_1rs^{n-1}$, …, $s | a_{n-1}r^{n-1}s$ and $s | 0$ we have that $s | a_nr^n$. Since $\gcd (r, s) = 1$ we must have that $s | a_n$.

- Furthermore, from above since $r | a_1rs^{n-1}$, …, $r | a_nr^n$, and $r | 0$ we have that $r | a_0s^n$. Since $\gcd (r, s) = 1$ we must have that $r | a_0$. $\blacksquare$