Degree of the polynomial
The polynomial P(X).
Choose a degree and enter the integer coefficients. At least one non-zero coefficient is needed.
If your polynonomial has fractional coefficients, go check the section Example 2: Can you use the rational zero test?

Welcome to the rational zeros calculator! It helps you perform the rational root test, that is, listing all possible rational zeros of an integer-coefficient polynomial. The calculator does it with help of the rational root theorem to accurately find the rational zeros of your polynomial.

If you aren't sure what finding rational zeros is all about, don't worry. Scroll down, we'll tell you what a rational zero is, what the rational zero theorem says, and how to find all possible rational zeros of a given polynomial. Then, we'll explain how to decide if a possible rational root is an actual root. In just a few paragraphs you'll be an expert on the rational zero test!

Are you wondering how to find the rational roots of just a good old quadratic equation? If so, then our parabola calculator will be perfect for you!

What is a rational zero?

Let us consider a polynomial in standard form with real coefficients (we assume an ≠ 0):

p(x) = anxn + an-1xn-1 + ... + a1x + a0.

  • We say that a real number, r, is a zero (or a root) of p if p evaluated at b gives zero: p(b) = 0; and

  • If r is a rational number (i.e., you can write it as p/q, where p and q are integers), then we call it a rational zero (or a rational root).

We now know what a rational zero is, so it's time to learn how to find rational roots! The first step is to find all possible rational roots, which we can do with help of the rational root theorem.

Rational root theorem (rational zero theorem)

Let p be as above. Assume that the coefficients an, an-1,..., a1, a0 of p are integers.

  • a0 is called the constant term of p (or the trailing coefficient); and
  • an is called the leading coefficient of p.

The rational root theorem says that if p has a rational root, then this root is equal to a fraction such that the numerator is a factor of a0 and the denominator is a factor of an (both positive and negative factors). In other words, every rational root of p fulfills the following:

± factor of a0 / factor of an

⚠️ The rational root theorem does not confirm that any of these numbers are actually zeros of our polynomial! It may happen that none of them are! What this theorem says is that IF there are any rational roots, THEN they must be of the form given above.

If an = 1, then we say that p is a monic polynomial. Note that in such case all possible rational roots of p are the factors of the trailing coefficient a0.

To summarize, the rational root theorem gives you the list of all possible rational zeros. A further rational root test allows you to determine all actual rational zeros. However, finding all roots of your polynomial may require additional work: remember that it may have irrational or even complex zeros as well!

How to find all possible rational zeros?

You can do it using the rational zero theorem which we described above:

  1. List all factors of the constant term. You can use our factor calculator.

  2. List all factors of the leading coefficient.

  3. To list all possible rational zeros, write down all possible fractions with the numerator taken from step 1 and denominator from step 2.

  4. Simplify the fractions and remove any duplicates.

❗ Remember: this is a list of all the possible rational roots of our polynomial. Each number from the list can (but doesn't have to) be a root. Any rational number not included in the list cannot be a root.

How to use this rational zeros calculator?

As you can see, finding rational zeros can be time-consuming: there might be lots of possible rational roots, and for each of them you have to check whether or not it's an actual zero. Fortunately, there's our rational zeros calculator, which can do all this work for you! 😊

Here's three simple steps which will show you how to find rational zeros with help of this rational zeros calculator:

  1. Choose the degree of your polynomial.

  2. Enter the polynomial's coefficients. Remember they must be integers!

    If your polynomial has some fractions as coefficients, multiply the whole polynomial by the least common denominator of these fractions. This new polynomial will have the same roots as your original polynomial.

  3. Our rational zeros calculator immediately displays the results - we list both possible rational zeros and actual rational zeros! 🎉

Example 1: How to find possible rational zeros?

Suppose we have a polynomial:

p(x) = 2x4 + 3x3 - 8x2 - 9x + 6

  • The factors of the constant term, 6:

    1, 2, 3, 6

  • The factors of the leading coefficient, 2:

    1 and 2

  • List of all possible rational zeros of p:

    ±1/1, ±1/2, ±2/1, ±2/2, ±3/1, ±3/2, ±6/1, ±6/2

  • Possible rational zeros of p after simplification and duplication removal:

    ±1, ±1/2, ±2, ±3, ±3/2, ±6

These twelve numbers (positive and negative) are all possible rational roots of p: no other rational number can be a root of p.

Example 2: Can you use the rational zero test?

Suppose we have the polynomial:

s(x) = (1/3)x3 + (3/4)x2 - 5x + 1/2

As you can see, s has non-integer coefficients. This means we can't use the rational root theorem to determine possible rational zeros of s! 😭

❗ Remember: rational root theorem applies only if all coefficients of the polynomial are integers.

So, what can we do? How to find rational zeros of a rational-coefficient polynomial?

What we need to do is to look at the fractions appearing in the polynomial and determine their least common denominator (LCD). In other words, we need to find the least common multiple (LCM) of the denominators of these fractions, provided that the fractions are in the simplest form. Then we multiply our polynomial by this LCD to get a polynomial with integer coefficients:

  • Fractional coefficients of s: 1/3, 3/4, 1/2.

  • Their LCD (the LCM of 3, 4, 2) is equal to 12.

  • Multiplying s by 12 gives the polynomial

4x3 + 9x2 - 60x + 6.

Surely, (1/3)x3 + (3/4)x2 - 5x + 1/2 and 4x3 + 9x2 - 60x + 6 are different polynomials, but they only differ by a constant factor, and so their roots coincide. The latter polynomial has integer coefficients, so we can apply the rational zero test!

  • The factors of the constant term, which is equal to 6:

    1, 2, 3, 6.

  • The factors of the leading coefficient, which is equal to 4:

    1, 2 and 4.

  • The list of all possible rational zeroes of s:

    ±1/1, ±1/2, ±1/4, ±2/1, ±2/2, ±2/4, ±3/1, ±3/2, ±3/4, ±6/1, ±6/2, ±6/4.

  • Possible rational zeroes of s after simplification:

±1, ±1/2, ±1/4, ±2, ±3, ±3/2, ±3/4, ±6.

How to find actual rational zeros?

In this section, we learn a bit about actually finding rational zeros. As we've learned above, the rational root theorem gives you all the potential rational roots, but doesn't guarantee that any of them are actually a root.

So, how to find rational roots given the list of all the potential rational roots? The simplest and most obvious answer is to take a possible rational root, substitute it for x in the polynomial, and check if the result is equal to zero. If so, then we have an actual root. If not, we discard the candidate and proceed to the next one, until we run out of possible rational roots. This straightforward, brute-force method leads to correct results, but can lure you into lengthy and complicated computations. 😩😰

Fear not! There is a better way and we'll explain it to you right now.

This better way is called polynomial division. What does it mean to divide polynomials? We can divide polynomials analogously to integers: we calculate quotients and remainders. Namely, let's divide a polynomial P(x) (dividend) by a non-zero polynomial D(x) (divisor). There exists a unique pair of polynomials Q(x) and R(x) which satisfy

P(x) = D(x) ⋅ Q(x) + R(x)

and deg(R) < deg(D). We call Q(x) the quotient and R(x) the remainder.

Maybe you've already learned how to divide polynomials using polynomial long division, but maybe you didn't know that there is a shortcut way - polynomial synthetic division. These methods only differ in the way things are written down: synthetic division is faster and more compact.

So, how to use polynomial division to verify if a candidate for a rational root actually nullifies the polynomial? Let r be our potential root. You just need to divide p(x) by x - r and look at the remainder of this division:

  • If the remainder is zero, then r is a root of p; and
  • If the remainder is non-zero, r is not a root of p.
💡 Tip. If a given candidate, r, is actually a root of p(x), you can use the quotient produced by the division p(x)/(x - r) to verify the remaining roots. This quotient is just the original polynomial from which the binomial, x - r, corresponding to the root has been factored out.

Rational root test: example

Let's see how to find rational zero of the polynomial:

p(x) = 2x4 + 3x3 - 8x2 - 9x + 6

We've already determined that its possible rational roots are ±1/2, ±1, ±2, ±3, ±3/2, ±6. Now we'll check which of them are actual rational zeros of p. Recall that r is a root of p if and only if the remainder from the division of p by x − r is equal to 0.

  • To check if 1/2 is a root of p, let's divide p by x − 1/2. The remainder is 0. So, 1/2 is a root of p. The quotient is equal to 2x3 + 4x2 - 6x - 12. In what follows we denote it by q. See the synthetic division calculator for details.

  • To check -1/2, let's divide q by x - 1/2.The remainder is 8.25. So, -1/2 is not a root of p.

  • For 1, divide q by x − 1.The remainder is -6. So, 1 is not a root of p.

  • For -1, divide q by x + 1.The remainder is 6. So, -1/2 is not a root of p.

  • For 2, divide q by x − 2.The remainder is 12. So, 2 is not a root of p.

  • For -2, divide q by x + 2.The remainder is 0. So, -2 is a root of p. The quotient is equal to 2x2 - 6.

In the same way we check the remaining six possible rational roots of p. It turns out that none of them are an actual root of p.

In fact, we can easily see that the roots of 2x2 - 6 are ±√3, and so p has two rational roots and two irrational roots.

Anna Szczepanek, PhD
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