Find Polynomial Roots With the Rational Root Theorem

Many students assume the rational root theorem only lists possibilities, but the real value is combining it with synthetic division to actually find the roots. The rational root theorem states that for a polynomial with integer coefficients, every rational zero is of the form p/q where p divides the constant term and q divides the leading coefficient. OpenStax College Algebra 2e section 5.5 covers this as part of Zeros of Polynomial Functions. Use this theorem to generate a short list of candidates, then test each one with synthetic division until you hit a remainder of zero. That zero confirms a factor, and the depressed quotient from the bottom row gives you a lower-degree polynomial to finish with the quadratic formula. The entire process turns a fifth-degree polynomial into a manageable problem in under ten minutes.

The rational zeros theorem gives you the p/q test: list all factors of the constant term over all factors of the leading coefficient. For a polynomial like 2x³ - 5x² + 4x - 3, the constant term is -3 and the leading coefficient is 2. Factors of -3: ±1, ±3. Factors of 2: ±1, ±2. That gives candidates ±1, ±3, ±1/2, ±3/2, eight possibilities to test. For a polynomial with large coefficients, this list can be long, so use Descartes' Rule of Signs first to trim it. OpenStax College Algebra 2e section 5.5 explains that the number of positive real roots is at most the number of sign changes in the coefficients, ignoring zeros. For 2x³ - 5x² + 4x - 3, the signs go +, -, +, -: three sign changes, meaning 3 or 1 positive roots. For negative roots, substitute -x and count again. This narrows the candidate list before you run a single synthetic division.

Set Up the Table Correctly

Set up the synthetic division table with the coefficients in a row: for 2x³ - 5x² + 4x - 3, write 2, -5, 4, -3. Choose a candidate, say 1, and place it to the left. Bring down the 2. Multiply by 1, add to -5 to get -3. Multiply -3 by 1, add to 4 to get 1. Multiply 1 by 1, add to -3 to get -2. The bottom row is 2, -3, 1, -2. The last number, -2, is the remainder. It is not zero, so x - 1 is not a factor. The most common error is the sign of c: dividing by x + 3 means c = -3, not 3. The remainder theorem tells you that P(1) = -2, which matches the remainder. A remainder of zero means x - c is a factor, and the numbers before the remainder are the quotient coefficients, degree one less than the dividend.

Work Through the Candidates

Test candidate -3: coefficients 2, -5, 4, -3 with c = -3 gives bottom row 2, -11, 37, -114, remainder not zero. Keep testing until one works. For this polynomial, try 3/2: with c = 1.5, the bottom row is 2, -2, 1, -1.5, not zero. The candidate 1/2 gives 2, -4, 2, -2, not zero. Try -1/2: c = -0.5 gives 2, -6, 7, -6.5, no. When none of the rational candidates work, the polynomial may have irrational or complex roots, but for many textbook problems, at least one rational root exists.

Once a candidate gives a remainder of zero, the bottom row excluding the last number gives the depressed quotient. For example, test x = 1 on 2x³ - 5x² + 4x - 3 gave remainder -2, so no. But test x = -1 on a different cubic: for x³ - 6x² + 11x - 6, coefficients 1, -6, 11, -6 with c = -1 gives bottom row 1, -7, 18, -24, remainder -24. Try x = 2: c = 2 gives 1, -4, 3, 0, remainder zero. The bottom row excluding the last is 1, -4, 3. That is the quotient: 1x² - 4x + 3. Set that equal to zero: x² - 4x + 3 = 0. Factor it or use the quadratic formula: x = (4 ± √(16 - 12))/2 = (4 ± 2)/2, giving x = 3 and x = 1. The full factorisation is (x - 2)(x - 3)(x - 1). The depressed quotient always has degree one less than the original polynomial, so a cubic reduces to a quadratic, which you solve by factoring or the quadratic formula. The remainder theorem confirms P(2) = 0, which matches the factor theorem: x - 2 is a factor because the remainder is zero.

Factor 2x³ + x² - 5x + 2 completely. The constant term is 2, leading coefficient is 2. Factors of 2: ±1, ±2. Factors of 2: ±1, ±2. Candidates: ±1, ±2, ±1/2. Descartes' Rule of Signs: signs +, +, -, +: two sign changes, so 2 or 0 positive roots. For negative roots, substitute -x: 2(-x)³ + (-x)² - 5(-x) + 2 = -2x³ + x² + 5x + 2: signs -, +, +, +: one sign change, so 1 negative root. Test candidates: try x = 1 with c = 1 on coefficients 2, 1, -5, 2. Bring down 2. Multiply by 1, add to 1 gives 3. Multiply 3 by 1, add to -5 gives -2. Multiply -2 by 1, add to 2 gives 0. Remainder zero. The bottom row excluding last is 2, 3, -2: quotient is 2x² + 3x - 2. Solve 2x² + 3x - 2 = 0 using the quadratic formula: x = [-3 ± √(9 + 16)] / (4) = [-3 ± 5] / 4. That gives x = 0.5 and x = -2. The full factorisation is (x - 1)(2x - 1)(x + 2). Check: multiply (x - 1)(2x - 1) = 2x² - 3x + 1, then multiply by (x + 2) gives 2x³ + x² - 5x + 2. The synthetic division higher-degree polynomials technique works the same for any degree, always bring down, multiply, add, repeat.

Factor x⁴ - 5x³ + 6x² + 4x - 8. Constant term -8, leading coefficient 1. Candidates are all factors of -8: ±1, ±2, ±4, ±8. Descartes' Rule of Signs: signs +, -, +, +, -: three sign changes, so 3 or 1 positive roots. Substitute -x: x⁴ + 5x³ + 6x² - 4x - 8: signs +, +, +, -, -: one sign change, so 1 negative root. Test candidates: try x = 2 with c = 2 on coefficients 1, -5, 6, 4, -8. Bring down 1. Multiply by 2, add to -5 gives -3. Multiply -3 by 2, add to 6 gives 0. Multiply 0 by 2, add to 4 gives 4. Multiply 4 by 2, add to -8 gives 0. Remainder zero. The bottom row excluding last is 1, -3, 0, 4: quotient is x³ - 3x² + 0x + 4. Now test the same candidate again on the depressed quotient: coefficients 1, -3, 0, 4 with c = 2. Bring down 1. Multiply by 2, add to -3 gives -1. Multiply -1 by 2, add to 0 gives -2. Multiply -2 by 2, add to 4 gives 0. Remainder zero again. The quotient is 1, -1, -2: x² - x - 2. Solve x² - x - 2 = 0: (x - 2)(x + 1) = 0, giving x = 2 and x = -1. The full factorisation is (x - 2)³(x + 1). Double-check: multiply (x - 2)³ = x³ - 6x² + 12x - 8, times (x + 1) gives x⁴ - 5x³ + 6x² + 4x - 8. The double root at x = 2 appears because the depressed quotient was divisible by x - 2 again. The candidate-list table for this quartic had eight entries, but only two needed testing before the polynomial broke apart.

Apply Descartes' Rule First

Before you test every candidate, use Descartes' Rule of Signs to cut the list in half. For the quartic above, the rule predicted 3 or 1 positive roots and 1 negative root. That meant you only needed to test negative candidates for the first root, and the positive candidates could wait. The candidate -1 worked, which matched the prediction.

Use a Graph to Spot Crossings

Graphing the polynomial on a calculator or an online tool shows where the curve crosses the x-axis. A graph of x⁴ - 5x³ + 6x² + 4x - 8 crosses at x = -1 and touches tangentially at x = 2, confirming the double root. Use the graph to pick which candidates to test first: the ones nearest the visible crossings. OpenStax College Algebra 2e section 5.5 covers Descartes' Rule of Signs in detail, including the rule that zero coefficients are ignored when counting sign changes. For a polynomial like x³ - 1 (which is x³ + 0x² + 0x - 1), the signs are +, -, so one sign change, meaning one positive root. The negative root count from f(-x) = -x³ - 1: signs -, -, so zero sign changes, zero negative roots. The only real root is 1. The rational root theorem gives candidate factors of 1 over factors of 1: just ±1. Test x = 1: remainder zero.The graph would show a single crossing at x = 1. The combination of Descartes' rule and a quick graph reduces a list of eight candidates to two or three that actually need testing.