Rational Root Test

What it is for

There is no quick formula for solving a polynomial of degree 3 or higher, so the usual approach is to find one root, factor it out, and repeat. The trouble is knowing where to start looking. The Rational Root Test turns an endless guessing game into a short checklist: it gives you the only whole numbers and fractions that could possibly be roots.

The rule

Take a polynomial whose coefficients are all integers. If a fraction p/q (in lowest terms) is a root, then:

• p must be a factor of the constant term, and
• q must be a factor of the leading coefficient.

So you list the factors of each, build every possible p/q (positive and negative), and test them one at a time.

A worked example

Find the roots of 2x³ + 5x² − 4x − 3.

1. Factors of the constant term −3 (p): ±1, ±3.
2. Factors of the leading coefficient 2 (q): ±1, ±2.
3. Possible roots p/q: ±1, ±3, ±1/2, ±3/2. That is only eight numbers to check.
4. Test them by substituting. Trying x = 1 gives 2 + 5 − 4 − 3 = 0, so 1 is a root and (x − 1) is a factor.
5. Divide by (x − 1) to get 2x² + 7x + 3, which factors as (2x + 1)(x + 3).
6. The roots are 1, −1/2 and −3. All three were on the list.

Try this example

Why it works

Suppose p/q is a root of ax³ + bx² + cx + d. Substitute it in and multiply both sides by q³ to clear the fractions:

ap³ + bp²q + cpq² + dq³ = 0

Every term except the last contains p, so p must divide dq³ as well. Since p and q share no factors, p must divide d, the constant term. In the same way, every term except the first contains q, so q must divide a, the leading coefficient. The same argument works for any degree.

What it does not tell you

• The list is of possible roots. Most candidates usually fail, and sometimes all of them do.
• It only finds rational roots. For x² − 2 the candidates are ±1 and ±2, and none work, because the real roots are ±√2, which are irrational.
• It needs integer coefficients. If yours are fractions, multiply the whole equation by a common denominator first.