The Rational Root Theorem says if a polynomial equation $ a_n x^n + a_{n – 1} x^{n – 1} + … + a_1 x + a_0 = 0$ has rational root $\frac{p}{q} (p, q \in \mathbb{Z})$ then the denominator q divides the leading coefficient and the numerator p divides $ a_0$.
As an addition to this theorem, for every whole number $k$, number $p – k \cdot q$ is a divisor of $f(k)$.
Example 1. Find all rational roots of the following equation:
The leading coefficient is 5 which means that, since q divides it, is from the set {-1, 1, -5, 5} and the free coefficient is number 3 which means that p is from the set {-1, 1, -3, 3}.
Since we know possibilities for q and p, we can find all combinations to see what our solutions, $\frac{p}{q}$ can be.
$\frac{p}{q} \in \{-1, 1, -\frac{1}{5}, \frac{1}{5}, -3, 3, -\frac{3}{5}, \frac{3}{5} \} $
The first solution is 1. When we factorize given equation we get:
$ 5x^3 – 7x^2 – x + 3 = (x – 1)(5x^2 – 2x – 3)$
Since the other factor is a quadratic polynomial, we can easily find its roots. As the final result we get:
$ x_1 = 1, x_2 = 1, x_3 = -\frac{3}{5}$
