The factor theorem turns a zero into a factor, but first you need a zero. For a polynomial with whole-number coefficients, the rational root theorem gives a short list of the only fractions worth trying.
Key idea
If $p(x) = a_nx^n + \dots + a_0$ has integer coefficients and $\frac{p}{q}$ (in lowest terms) is a zero, then $p$ divides the constant $a_0$ and $q$ divides the leading coefficient $a_n$. So the candidates are $\pm\frac{\text{factor of } a_0}{\text{factor of } a_n}$. Test them, and divide out each zero you find.
Why it works
Suppose $\frac{p}{q}$ is a zero of $2x^3 + \dots + a_0$. Substitute it and multiply through by $q^3$: $$2p^3 + (\text{terms with } q) + a_0q^3 = 0.$$ Every term except $a_0q^3$ has a factor of $p$, so $p$ divides $a_0q^3$. Since $p$ and $q$ share no factors, $p$ divides $a_0$. Looking at the first term the same way, $q$ divides $2p^3$ and therefore divides 2, the leading coefficient.
Worked examples
Worked example: The list of candidates
List the possible rational zeros of $p(x) = 2x^3 - 3x^2 - 3x + 2$.
- Factors of the constant 2: $1, 2$. Factors of the leading coefficient 2: $1, 2$.
- $\pm\frac{1}{1}, \pm\frac{2}{1}, \pm\frac{1}{2}, \pm\frac{2}{2}$.
Answer: $\pm 1, \pm 2, \pm\frac{1}{2}$
Worked example: Finding all the rational zeros
Find all rational zeros of the same $p(x)$.
- Try $x = 2$: $16 - 12 - 6 + 2 = 0$. So $x - 2$ is a factor.
- Synthetic division by 2 gives $2x^2 + x - 1$.
- Factor: $2x^2 + x - 1 = (2x - 1)(x + 1)$.
Answer: $2$, $\frac{1}{2}$ and $-1$
Common mistake
Swapping $p$ and $q$: putting factors of the leading coefficient on top. For $2x^3 - 3x^2 - 3x + 9$, $\frac{9}{2}$ is a candidate but $\frac{2}{9}$ is not. Remember "constant over leading." Also remember that the list only gives candidates; a candidate is a zero only if it tests to 0.