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Rational Root Theorem Practice Problems

The rational root theorem does not find roots; it produces a finite shortlist to test. These twelve problems drill both halves of that — building the candidate list correctly from the first and last coefficients, and then narrowing it down to the roots that are actually there.

  1. 1.

    List every possible rational root of .

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    The constant term is , so divides 6: . The leading coefficient is , so divides 1 and the only option is . Every candidate is therefore a factor of 6: .

  2. 2.

    How many possible rational roots does the theorem list for ?

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    Factors of the constant 6 give , and the leading coefficient of 1 gives . That is four values, each with two signs, so 8 candidates.

  3. 3.

    Find all rational roots of .

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    Candidates are . Testing: and . Both work, which matches the factorisation .

  4. 4.

    For , list the possible values of — the denominators in .

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    runs over the divisors of the leading coefficient, which is 2. Its divisors are 1 and 2, so . The constant term plays no part in choosing .

  5. 5.

    List every possible rational root of .

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    divides 12: . divides 2: . Over you get . Over , the halves duplicate values already listed, leaving only and as new. Sixteen candidates in total.

  6. 6.

    Find all rational roots of .

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    Test : , so it is a root. Divide synthetically by to get the quotient , which factors as . That gives and . All three roots are rational: .

  7. 7.

    Find all rational roots of .

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    Candidates are . Test : . Dividing by leaves , giving and .

  8. 8.

    How many possible rational roots does the theorem list for ?

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    divides 10: . divides 3: . The distinct quotients are — eight values, none duplicated — so 16 with signs.

  9. 9.

    Find all rational roots of .

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    Test : . Synthetic division by gives the quotient , which factors as . The roots are .

  10. 10.

    Find all rational roots of .

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    With and the shortlist is long, so start with the whole numbers. . Dividing by leaves , giving and . Testing integers before fractions is almost always the faster order.

  11. 11.

    Given that is a root of , find the other two roots.

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    Divide by synthetically: the bottom row is with remainder 0, so the quotient is . The other roots are and .

  12. 12.

    Find all rational roots of .

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    , so is a root. Dividing by gives , a difference of squares equal to . The remaining roots are .
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MPhil in Mathematics, Founder of Mathovia
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Rational Root Theorem Practice Problems

The rational root theorem lesson explains where the rule comes from. These twelve problems put it to work, and they are graded so that the early ones only build the candidate list while the later ones follow through to the actual roots.

What the theorem gives you

is the constant term and the leading coefficient. Note which is which — swapping them is the single most common error, and it produces a list that usually contains none of the real roots.

How to work through a problem

  1. Read off and . Write the divisors of each.
  2. Form every distinct , with both signs. Simplify, and drop duplicates.
  3. Test the small integers first. Synthetic division is quickest, and a hit reduces the degree.
  4. Factor what is left. Once the polynomial drops to a quadratic, factor or use the quadratic formula instead of continuing to test.

The last step matters. There is no reason to keep testing candidates once the remaining factor is a quadratic — factoring it is faster and finds irrational roots too, which the theorem never will.

Frequently Asked Questions

What does the rational root theorem actually tell you?+

That if a polynomial with integer coefficients has a rational root p/q in lowest terms, then p divides the constant term and q divides the leading coefficient. It produces a finite shortlist of candidates — it does not tell you which of them, if any, are real roots.

Does the theorem guarantee a rational root exists?+

No. A polynomial can have a full candidate list and no rational roots at all. x² − 2 lists ±1, ±2 as candidates and none of them work, because its roots are the irrational ±√2.

Which candidates should I test first?+

The small integers, especially ±1 and ±2. They are quickest to evaluate mentally, and they are disproportionately often the actual roots in textbook polynomials. Move to the fractions only when the whole numbers are exhausted.

What is the fastest way to test a candidate?+

Synthetic division. It evaluates P(a) and, when the remainder is zero, hands you the depressed polynomial in the same pass — so a successful test immediately reduces the degree you still have to solve.

Why do some candidate lists have duplicates?+

Because p/q can simplify. For a leading coefficient of 2, the candidate 4/2 is just 2, which is already on the list from q = 1. List the distinct values, not the raw pairs, or the count comes out too high.

Does it work if the coefficients are fractions?+

Not directly — the theorem needs integer coefficients. Multiply the whole polynomial through by the least common denominator first. That does not change the roots, and the result is an integer-coefficient polynomial the theorem applies to.

Calculators for this topic

Zeros Calculator

More practice problems

Polynomial Zeros Practice Problems Synthetic Division Practice Problems Remainder and Factor Theorem Practice Problems