Lesson · Polynomials and Factoring · Division and roots

The Factor Theorem and Rational Roots

The factor theorem converts a question about values into a question about factors. The rational root theorem then supplies a finite list of candidates, so factoring a cubic stops being inspiration and becomes a search that can be completed — including when it finds nothing.

Open the interactive lesson Practice

f(c) = 0 ⟺ (x - c) is a factor of f root p/q in lowest terms ⟹ p | a₀ and q | aₙ 2x³ - 3x² - 8x - 3 = (x + 1)(2x + 1)(x - 3)

Two theorems, one search

Core ideas

What this lesson establishes

The first theorem says what a root is worth. The second says where to look. Together they make the search finite.

A root and a factor are the same information

f(c) = 0 and (x - c) divides f are equivalent statements, not two related facts. Each direction follows from the remainder theorem in one line.

The candidate list is generated, not guessed

Every rational root has numerator dividing the constant term and denominator dividing the leading coefficient. That gives a finite list you can write down before testing anything.

An empty list is a conclusion

If no candidate is a root, the polynomial has no rational roots at all. It may still have irrational or complex ones — x³ - 2 does — but none that this course can produce exactly, and saying so is the correct answer.

The material

From roots to factors, and back

Two theorems, a proof each, and the search they license.

Theorem · The factor theorem

For a polynomial f and a number c: (x - c) is a factor of f if and only if f(c) = 0.

By the remainder theorem, f(x) = (x - c)q(x) + f(c).

If f(c) = 0 this reads f(x) = (x - c)q(x), so (x - c) is a factor.

Conversely, if (x - c) is a factor then f(x) = (x - c)h(x) for some polynomial h, and substituting x = c gives f(c) = 0.

□

Both directions are used. Left to right, a root you have found hands you a factor and lets you divide the degree down by one. Right to left, a factor you have found tells you a root without any further work.

Where to look for roots

Theorem · The rational root theorem

Let f(x) = aₙxⁿ + … + a₁x + a₀ have integer coefficients with aₙ ≠ 0 and a₀ ≠ 0. If the rational number p/q, in lowest terms, is a root of f, then p divides a₀ and q divides aₙ.

Read the direction carefully. The theorem does not say that every such p/q is a root — almost all of them are not. It says that any rational root must be on that list, which is what makes checking the list a complete search.

For 2x³ - 3x² - 8x - 3 the constant term is -3, whose divisors are ±1 and ±3; the leading coefficient is 2, whose divisors are ±1 and ±2. Every candidate p/q is therefore one of ±1, ±3, ±1/2, ±3/2 — eight numbers, and any rational root is among them.

Example · Factoring 2x³ - 3x² - 8x - 3 completely

Test c = -1: f(-1) = -2 - 3 + 8 - 3 = 0, so (x + 1) is a factor. Synthetic division by x + 1 gives the quotient 2x² - 5x - 3, which the AC method factors as (2x + 1)(x - 3). Hence 2x³ - 3x² - 8x - 3 = (x + 1)(2x + 1)(x - 3), with roots -1, -1/2 and 3.

Notice what happened after the first success. The problem became a quadratic, and quadratics are already solved — so the candidate list is only needed until the degree drops to two. Testing more candidates against the original cubic would be work with nothing to show for it.

When the search comes back empty

Example · x³ - 2 has no rational root

The constant term is -2 and the leading coefficient is 1, so the candidates are 1, -1, 2, -2. Evaluating: f(1) = -1, f(-1) = -3, f(2) = 6, f(-2) = -10. None is zero, and the list is exhausted. x³ - 2 has no rational root, and no rational linear factor.

It does have a real root, ∛2, which is irrational; the theorem is exactly the tool that proves ∛2 cannot be written as a fraction. This course states that the root is irrational and stops there. Approximating it is a numerical question, not an algebraic one.

Theorem · How many roots there can be

A polynomial of degree n has at most n roots. Each root c contributes a factor (x - c), and the degrees of the factors cannot exceed n.

Interactive

The rational root theorem

Candidates generated from the divisors, then tested

The whole candidate list, tested

Enter the coefficients. The lab lists every divisor of a₀ and of aₙ, forms every candidate p/q, and evaluates each in exact fractions. Try x³ - 2 and watch the list finish with no root found.

Example

The polynomial

Candidates—

Roots found—

Factored form—

Multiplies back—

Type as you would write it: 6x^3 + 9x^2 - 15x, (1/2)x^2 + 3/4. Every divisor list, every pair tried and every test below is computed in your browser from what is in this box.

Method

Factoring a polynomial of degree 3 or more

The list first, then one root at a time, then stop early.

Take out the common factor

A common factor lowers the constant and leading coefficients, and so shortens the candidate list before it is written. 2x³ - 6x² + 4x becomes 2x(x² - 3x + 2), and the hard part is now a quadratic.

Build the candidate list

List the divisors of the constant term as p, the divisors of the leading coefficient as q, and form every ± p/q. Remove duplicates; 2/2 and 1/1 are the same candidate.

Test candidates until one is a root

Evaluate f at each, by synthetic division so that a hit hands you the quotient immediately. f(c) = 0 means (x - c) is a factor.

Repeat on the quotient, and stop at degree 2

The quotient is one degree lower; search it the same way. Once it is a quadratic, use the pair or ac method already taught. If those searches are empty, record that no rational linear factor remains; Quadratics and Complex Numbers classifies its irrational or complex roots.

Check yourself

Score 0 / 0

Choose the answer you can justify.

Factoring 2x³ - 3x² - 8x - 3

The constant term is -3 and the leading coefficient is 2, so the candidate list has eight entries and any rational root is among them.

f(x) = 2x³ - 3x² - 8x - 3 a₀ = -3, aₙ = 2 p divides -3 : ±1, ±3 q divides 2 : ±1, ±2 candidates p/q: ±1, ±3, ±1/2, ±3/2 (8 in total) f(1) = 2 - 3 - 8 - 3 = -12 not a root f(-1) = -2 - 3 + 8 - 3 = 0 ROOT -> (x + 1) is a factor synthetic division by x + 1, c = -1 : -1 │ 2 -3 -8 -3 │ -2 5 3 └──────────────────────── 2 -5 -3 0 remainder 0, as required quotient 2x² - 5x - 3 AC method: ac = -6, pair -6 and 1, sum -5 2x² - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3) 2x³ - 3x² - 8x - 3 = (x + 1)(2x + 1)(x - 3) roots: x = -1, x = -1/2, x = 3

The search stopped after two candidates, and that is normal — but it stopped for a reason, not from fatigue. Once (x + 1) came out, the remaining problem was a quadratic, and the candidate list had done its job.

Every root that appeared is on the original list: -1, -1/2 and 3 are all of the form p/q with p dividing -3 and q dividing 2. That is the theorem doing what it promised, and it is worth checking once to believe it.

For a faded rational-root search, take 2x³ - 3x² - 5x + 6. Build the candidate list, with the supplied first test f(1) = 0; divide by x - 1, then factor the quotient by the ac method. The complete result is (x - 1)(2x + 3)(x - 2). If your quotient is 2x² + x - 6, the synthetic addition under the -3 coefficient used the wrong sign.

Common mistakes

Reading the theorem as a promise that candidates are rootsThe list is where roots must be, not a list of roots. For a typical cubic most of the eight candidates fail, and that is the theorem working correctly.

Getting p and q the wrong way roundThe numerator divides the CONSTANT term and the denominator divides the LEADING coefficient. Swapping them for 2x³ - 3x² - 8x - 3 would offer ±1, ±2, ±1/3, ±2/3 — eight candidates, of which only -1 happens to be a root — and miss the actual root -1/2 entirely.

Continuing to test candidates after the degree drops to twoOnce the quotient is a quadratic, use the pair or ac method from “Factoring Simple Trinomials” and “The AC Method”. Further candidate testing can only rediscover what those methods give directly, and an empty rational search must not be inflated into a claim that the quadratic has no real or complex roots.

Check your understanding

✓

Finish when a complete search that finds nothing reads as a proof.

You should factor 2x³ - 3x² - 8x - 3 down to three linear factors, and state of x³ - 2 that it has no rational root — citing the four tested candidates as the reason rather than reporting that you could not find one.

Note: This is the lesson the course was arranged around. Rational and Radical Expressions needs factored denominators, Quadratics and Complex Numbers needs the roots of quadratics, and “Graphs of Polynomial Functions” needs the factored form to draw a graph. All three are downstream of a search that either produces exact factors or proves there are none.