Rational Root Theorem

The Problem

You have a polynomial and want to find its roots.

For quadratics, you have the quadratic formula. But for higher degrees, there’s no simple formula.

The Rational Root Theorem helps by telling you which rational numbers to try.


The Theorem

For a polynomial with integer coefficients:

anxn+an1xn1++a1x+a0a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0

If pq\frac{p}{q} is a rational root (in lowest terms), then:

  • pp divides the constant term a0a_0
  • qq divides the leading coefficient ana_n

Possible rational roots = ±factors of constantfactors of leading coefficient\pm \frac{\text{factors of constant}}{\text{factors of leading coefficient}}


Example

Find rational roots of: 2x33x28x+122x^3 - 3x^2 - 8x + 12


Step 1: Identify the key coefficients

  • Leading coefficient = 22
  • Constant term = 1212

Step 2: List factors

  • Factors of 1212: 1,2,3,4,6,121, 2, 3, 4, 6, 12
  • Factors of 22: 1,21, 2

Step 3: Form all possible rational roots

±1,2,3,4,6,121,2\pm \frac{1, 2, 3, 4, 6, 12}{1, 2}

This gives: ±1,±2,±3,±4,±6,±12,±12,±32\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}


Step 4: Test candidates

Try x=2x = 2:

2(2)33(2)28(2)+12=161216+12=0\begin{aligned} 2(2)^3 - 3(2)^2 - 8(2) + 12 &= 16 - 12 - 16 + 12 \\ &= 0 \end{aligned}

So x=2x = 2 is a root.


Important Note

The theorem gives you candidates, not guaranteed roots.

  • If a rational root exists, it’s in this list
  • But many candidates won’t actually be roots
  • Some polynomials have no rational roots (like x22x^2 - 2, whose roots are ±2\pm\sqrt{2})