To find all possible rational zeros of a polynomial, use the Rational Zero Theorem. It turns your polynomial into a finite “checklist” of candidate fractions that might be zeros, which you then test.

Core idea (Rational Zero Theorem)

For a polynomial with integer coefficients
f(x)=anxn+⋯+a1x+a0f(x)=a_nx^n+\dots +a_1x+a_0f(x)=an​xn+⋯+a1​x+a0​,

  • Any rational zero must have the form pq\frac{p}{q}qp​, where:
    • ppp is a factor of the constant term a0a_0a0​
    • qqq is a factor of the leading coefficient ana_nan​.
  • You must consider both positive and negative values of pq\frac{p}{q}qp​.
  • This gives all possible rational zeros , but not all of them will actually work when plugged into the polynomial.

So the job is: list all ±factor of a0factor of an\pm \frac{\text{factor of }a_0}{\text{factor of }a_n}±factor of an​factor of a0​​, simplify duplicates, then test.

Step-by-step recipe

Use this general 4-step process for any polynomial with integer coefficients:

  1. Identify coefficients
    • Write the polynomial clearly and pick out:
      • The leading coefficient (first term’s coefficient) = ana_nan​
      • The constant term (no xxx) = a0a_0a0​.
  1. List factors
    • List all integer factors of a0a_0a0​:
      • Example: if a0=−4a_0=-4a0​=−4, factors are ±1,±2,±4\pm 1,\pm 2,\pm 4±1,±2,±4.
 * List all integer factors of ana_nan​:
   * Example: if an=2a_n=2an​=2, factors are ±1,±2\pm 1,\pm 2±1,±2.
  1. Form all fractions p/qp/qp/q
    • Form every fraction pq\frac{p}{q}qp​ where:
      • ppp comes from factors of a0a_0a0​
      • qqq comes from factors of ana_nan​.
 * Include **both** plus and minus: ±pq\pm \frac{p}{q}±qp​.
 * Simplify any fractions and remove duplicates:
   * Example: 22=1\frac{2}{2}=122​=1, 42=2\frac{4}{2}=224​=2, so you don’t list 2 twice.
  1. Test which candidates are actual zeros
    • The list from step 3 is just “maybes.”
    • Plug each candidate into the polynomial:
      • If f(r)=0f(r)=0f(r)=0, then rrr is an actual rational zero.
    • Often synthetic division and the Factor Theorem are used here to speed things up.

Quick concrete example

Suppose
f(x)=2x4−5x3+x2−4f(x)=2x^4-5x^3+x^2-4f(x)=2x4−5x3+x2−4.

  1. Leading coefficient: an=2a_n=2an​=2.
    Constant term: a0=−4a_0=-4a0​=−4.
  1. Factors:
    • Factors of −4-4−4: ±1,±2,±4\pm 1,\pm 2,\pm 4±1,±2,±4 (these are the ppp’s).
 * Factors of 222: ±1,±2\pm 1,\pm 2±1,±2 (these are the qqq’s).
  1. Form pq\frac{p}{q}qp​ and simplify:

    • Raw combinations include:
      ±1,±2,±4,±12,±22,±42\pm 1,\pm 2,\pm 4,\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{4}{2}±1,±2,±4,±21​,±22​,±24​, etc.

    • After simplification and removing duplicates, the possible rational zeros are:
      ±1,±2,±4,±12\pm 1,\pm 2,\pm 4,\pm \frac{1}{2}±1,±2,±4,±21​.

  1. Next, you would test each of these values to see which ones actually make f(x)=0f(x)=0f(x)=0.

Special easy case

  • If the leading coefficient is 1 (a monic polynomial), then:
    • The possible rational zeros are just the integer factors of the constant term (with ±).
* Example: f(x)=x3−5x+6f(x)=x^3-5x+6f(x)=x3−5x+6 → constant term 6 → possible rational zeros: ±1,±2,±3,±6\pm 1,\pm 2,\pm 3,\pm 6±1,±2,±3,±6.

Important limitations

  • The method only gives rational candidates.
  • It does not list irrational or complex zeros (like 5\sqrt{5}5​ or 3+2i3+2i3+2i).
  • Some or many of the candidates will not be actual zeros; they are just the complete set of possibilities that you need to check.

If you have a specific polynomial, share it and the possible rational zeros can be listed explicitly using this process.