Rational Root Theorem

Math Algebra

Proof

Let polynomial

$$P(x)=\sum _{i=0}^n{c}_i{x}^i$$

where $c_i\in \mathbb{Z}$ (all values of $c$ are integers).

Now let $P(\frac{p}{q})=0$, where $p$ and $q$ are coprime integers (let a fraction $\frac{p}{q}$ be in simplest form and be a root of $P$).

$$\sum _{i=0}^n{c}_i(\frac{p}{q}{)}^i=0$$

Multiplying by $q^n$:

$$\sum _{i=0}^n{c}_i{p}^n{q}^{n-i}=0$$

Now subtract $c_0{q}^n$ from both sides and factor $p$ out to get:

$$p\sum _{i=1}^n{c}_i{p}^{n-1}{q}^{n-i}=-c_0{q}^n$$

Now $p$ must divide $-c_0{q}^n$. However, we know $p$ cannot divide $q^n$ (since $\frac{p}{q}$ is in simplest form / $p$ and $q$ are coprime), so $p$ must divide $c_0$.

Doing the same thing as above but with the $a_n$ term and $q$:

$$q\sum _{i=0}^{n-1}{c}_i{p}^n{q}^{n-i-1}=-c_n{p}^n$$

By the above logic, $q$ must divide $c_n$.

Conclusion

For all rational roots in simplest form ($\frac{p}{q}$ where $p$ and $q$ are coprime integers), $p$ must be a factor of the last coefficient while $q$ must be a factor of the first coefficient.

Notes

For the curious, coprime integers $p$ and $q$ mean that $\gcd (p,q)=1$.

If future me or someone else is wondering about the excess definitions, this was made for a friend.