Factor Theorem Explained: How to Test Polynomial Factors

The Factor Theorem states that x − a is a factor of a polynomial f(x) if and only if f(a) = 0. Testing whether x − 4 divides f(x) = x³ − 3x² − 6x + 8 takes one substitution: f(4) = 64 − 48 − 24 + 8 = 0, so x − 4 is a factor.

That is the whole test. No division, no guessing, no trial multiplication. A single evaluation decides the question, and the same evaluation tells you a root of the equation f(x) = 0. The Remainder Theorem Calculator runs the same substitution and reports a factor verdict alongside it.

This guide states the theorem in both directions, shows how to test a candidate factor, works through factoring cubics and quartics completely, covers questions where one or two coefficients are unknown, and finishes with practice problems and answers.

What the Factor Theorem Says

The theorem is an “if and only if” statement, which means it works in both directions.

x − a is a factor of f(x) ⟺ f(a) = 0

Reading it left to right: if x − a divides f(x) exactly, then substituting a into f(x) gives 0.

Reading it right to left: if substituting a into f(x) gives 0, then x − a divides f(x) exactly.

Both directions matter in practice. The first lets you predict a value. The second lets you prove a factor.

Factor, Root and Zero Mean the Same Thing

Three pieces of vocabulary describe one situation, and exam questions switch between them freely.

All four say f(a) = 0. If a question gives you any one of them, you have all four.

Note where the sign sits. The factor x − 4 corresponds to the root +4. The factor x + 2 corresponds to the root −2. Reversing that is the most common error in the topic.

Why the Theorem Is True

The Factor Theorem follows directly from the polynomial division identity. Dividing f(x) by x − a gives

f(x) = (x − a) · q(x) + r

where r is a constant, because a degree-1 divisor forces a degree-0 remainder. The Remainder Theorem identifies that constant: setting x = a collapses the first term to zero and leaves r = f(a).

Now suppose f(a) = 0. Then r = 0, and the identity becomes

f(x) = (x − a) · q(x)

which is exactly the statement that x − a is a factor.

Suppose instead that x − a is a factor. Then f(x) = (x − a)·q(x) with no remainder term, and substituting x = a gives f(a) = 0 · q(a) = 0.

Both directions are proved, so the Factor Theorem is simply the Remainder Theorem restricted to the single case r = 0.

How to Test Whether x − a Is a Factor

1. Set the candidate factor equal to zero and solve for x. The candidate x − 4 gives a = 4. The candidate x + 2 gives a = −2.
2. Substitute a into f(x).
3. Compare with zero. A result of 0 means it is a factor. Anything else means it is not, and that value is the remainder.

Example: A Factor

Is x + 2 a factor of f(x) = x³ + x² − 4x − 4?

Solve x + 2 = 0 to get a = −2.

f(−2) = (−2)³ + (−2)² − 4(−2) − 4
      = −8 + 4 + 8 − 4
      = 0

Yes. x + 2 is a factor, and −2 is a root of f(x) = 0.

Example: Not a Factor

Is x − 3 a factor of f(x) = 2x³ − x² + 4?

Solve x − 3 = 0 to get a = 3.

f(3) = 2(27) − 9 + 4
     = 54 − 9 + 4
     = 49

No. The value is 49, not 0, so x − 3 is not a factor. The extra information here is free: 49 is the remainder when f(x) is divided by x − 3, which is the same leftover polynomial long division would produce after three rounds of work.

Factoring a Cubic Completely

The Factor Theorem finds the first factor. After that, division reduces the problem to a quadratic you already know how to handle.

Problem. Factor f(x) = 2x³ − 3x² − 11x + 6 completely.

Step 1: List the candidates. A rational root p/q must have p dividing the constant term 6 and q dividing the leading coefficient 2. So the candidates are

±1, ±2, ±3, ±6, ±1/2, ±3/2

Step 2: Test until one works.

f(1) = 2 − 3 − 11 + 6 = −6      not a root
f(2) = 16 − 12 − 22 + 6 = −12   not a root
f(3) = 54 − 27 − 33 + 6 = 0     root found

So x − 3 is a factor.

Step 3: Divide it out. Synthetic division with a = 3 on the coefficients 2, −3, −11, 6:

The quotient is 2x² + 3x − 2, and the remainder 0 confirms the factor.

Step 4: Factor the quadratic. 2x² + 3x − 2 = (2x − 1)(x + 2).

Answer.

2x³ − 3x² − 11x + 6 = (x − 3)(2x − 1)(x + 2)

The roots are 3, 1/2 and −2.

Verification. Check one root that was not used to build the answer: f(−2) = 2(−8) − 3(4) − 11(−2) + 6 = −16 − 12 + 22 + 6 = 0. ✓

Factoring a Quartic

The same routine repeats. Each factor found drops the degree by one.

Problem. Factor f(x) = x⁴ − 5x² + 4 completely.

Test a = 1: f(1) = 1 − 5 + 4 = 0, so x − 1 is a factor.

Test a = −1: f(−1) = 1 − 5 + 4 = 0, so x + 1 is a factor.

Test a = 2: f(2) = 16 − 20 + 4 = 0, so x − 2 is a factor.

Test a = −2: f(−2) = 16 − 20 + 4 = 0, so x + 2 is a factor.

A quartic has at most four linear factors, and four have been found, so

x⁴ − 5x² + 4 = (x − 1)(x + 1)(x − 2)(x + 2)

Verification. (x − 1)(x + 1) = x² − 1 and (x − 2)(x + 2) = x² − 4. Multiplying: (x² − 1)(x² − 4) = x⁴ − 4x² − x² + 4 = x⁴ − 5x² + 4. ✓

Finding One Unknown Coefficient

If a question tells you that something is a factor, that is an equation in disguise.

Problem. Find k so that x − 2 is a factor of f(x) = x³ + kx² − 4x + 4.

x − 2 being a factor means f(2) = 0:

f(2) = 8 + 4k − 8 + 4 = 4 + 4k

4 + 4k = 0
     k = −1

Check. With k = −1 the polynomial is x³ − x² − 4x + 4, and f(2) = 8 − 4 − 8 + 4 = 0. ✓

Finding Two Unknown Coefficients

Two stated factors give two equations, so two unknowns can be recovered.

Problem. f(x) = x³ + ax² + bx − 6 has x − 1 and x + 3 as factors. Find a and b, then factor f(x) completely.

From f(1) = 0:

1 + a + b − 6 = 0     →     a + b = 5

From f(−3) = 0:

−27 + 9a − 3b − 6 = 0     →     9a − 3b = 33     →     3a − b = 11

Add the two simplified equations to eliminate b:

(a + b) + (3a − b) = 5 + 11
              4a   = 16
               a   = 4

Substituting back gives b = 1, so f(x) = x³ + 4x² + x − 6.

Find the third factor. The known factors multiply to (x − 1)(x + 3) = x² + 2x − 3, which is quadratic, so the missing factor is linear. Because the constant term of f(x) is −6 and (−1)(3) = −3, the remaining constant must be 2:

x³ + 4x² + x − 6 = (x − 1)(x + 3)(x + 2)

Check. f(−2) = −8 + 16 − 2 − 6 = 0. ✓ And f(1) = 1 + 4 + 1 − 6 = 0, f(−3) = −27 + 36 − 3 − 6 = 0. ✓

Factor Theorem vs Remainder Theorem

The two theorems answer different questions from the same calculation.

In practice you run one substitution and get both answers at once. If f(a) turns out to be 0, you have a factor. If it turns out to be 49, you have a remainder of 49 and no factor.

Common Factor Theorem Mistakes

• Flipping the sign. The factor x + 5 corresponds to a = −5. Solve the candidate for zero rather than reading the number off the page.
• Testing f(a) ≠ 0 and concluding nothing. A non-zero result is still useful: it is the remainder.
• Assuming every polynomial has a rational root. x³ + x + 1 has no rational root, so no candidate from the list will give 0. The theorem does not fail; the polynomial simply has no linear factor with rational coefficients.
• Forgetting the denominators in the candidate list. For 2x³ − 3x² − 11x + 6 the candidates include ±1/2 and ±3/2, because the leading coefficient is 2.
• Stopping after one factor. A cubic with a factor found still has a quadratic to factor, which may split further.
• Confusing a factor with a root. x − 3 is the factor; 3 is the root. Answers to “find the roots” should be numbers, not expressions.

Factor Theorem Practice Problems

1. Show that x − 2 is a factor of f(x) = x³ − 2x² − 9x + 18, then factor f(x) completely.
2. Show that x + 4 is a factor of f(x) = x³ + 5x² + 2x − 8, then factor f(x) completely.
3. Show that x − 1 is a factor of f(x) = 3x³ + 2x² − 7x + 2, then factor f(x) completely.
4. Determine whether x − 3 is a factor of f(x) = x⁴ − 3x³ + 2x − 8.
5. Factor f(x) = 2x³ + 9x² + 7x − 6 completely, given that −2 is a root.
6. Find k so that x + 2 is a factor of f(x) = x³ + kx² + 2x + 8.
7. Factor f(x) = x³ − 6x² + 11x − 6 completely.
8. f(x) = x³ + px² + qx + 12 has factors x − 2 and x − 3. Find p and q, and give the third root.

Answers

1. f(2) = 8 − 8 − 18 + 18 = 0. Synthetic division gives the quotient x² − 9, so f(x) = (x − 2)(x − 3)(x + 3).
2. f(−4) = −64 + 80 − 8 − 8 = 0. The quotient is x² + x − 2, so f(x) = (x + 4)(x + 2)(x − 1).
3. f(1) = 3 + 2 − 7 + 2 = 0. The quotient is 3x² + 5x − 2, so f(x) = (x − 1)(3x − 1)(x + 2).
4. f(3) = 81 − 81 + 6 − 8 = −2. Not zero, so x − 3 is not a factor. The remainder is −2.
5. f(−2) = −16 + 36 − 14 − 6 = 0. The quotient is 2x² + 5x − 3, so f(x) = (x + 2)(2x − 1)(x + 3).
6. f(−2) = −8 + 4k − 4 + 8 = 4k − 4, and 4k − 4 = 0 gives k = 1.
7. f(1) = 1 − 6 + 11 − 6 = 0, and the quotient is x² − 5x + 6, so f(x) = (x − 1)(x − 2)(x − 3).
8. f(2) = 8 + 4p + 2q + 12 = 0 gives 4p + 2q = −20, so 2p + q = −10. f(3) = 27 + 9p + 3q + 12 = 0 gives 9p + 3q = −39, so 3p + q = −13. Subtracting gives p = −3, then q = −4. So f(x) = x³ − 3x² − 4x + 12, and since the constant 12 equals 2 × 3 × (−2) with a sign change, the third factor is x + 2 and the third root is −2. Check: f(−2) = −8 − 12 + 8 + 12 = 0. ✓

Factor Theorem FAQ

What is the Factor Theorem in simple terms?

It says that plugging a number into a polynomial and getting 0 proves that the matching linear expression divides the polynomial exactly. Get 0 at x = 5 and x − 5 is a factor. Get anything else and it is not, and that value is the remainder, exactly as a remainder of zero signals an exact division for whole numbers.

How do you use the Factor Theorem to factor a cubic?

Test small candidate values until one gives 0, which produces the first factor. Divide it out with synthetic division to get a quadratic quotient, then factor that quadratic by inspection or with the quadratic formula. The three linear factors together are the full factorisation.

What is the difference between the Factor Theorem and the Remainder Theorem?

The Remainder Theorem gives the remainder f(a) for any value of a. The Factor Theorem is the special case where that remainder is 0. One substitution answers both questions, so in practice you use them together rather than choosing between them.

Does the Factor Theorem work for x + a?

Yes. Write x + a as x − (−a) and substitute the negative value. To test x + 7, evaluate f(−7). A result of 0 means x + 7 is a factor and −7 is a root.

What if no candidate value gives zero?

Then the polynomial has no rational root, and therefore no linear factor with rational coefficients. x² + 1 and x³ + x + 1 are examples. The polynomial may still factor over the real or complex numbers, but not into the kind of factor this theorem tests for.

How do I know which values to test?

Use the Rational Root Theorem. Any rational root p/q has p dividing the constant term and q dividing the leading coefficient. For x³ − 6x² + 11x − 6 the leading coefficient is 1 and the constant is −6, so the candidates are ±1, ±2, ±3 and ±6.

Can x − a be a factor more than once?

Yes. If f(a) = 0 and the quotient also gives 0 at a, then (x − a)² divides f(x), which is called a repeated root. For f(x) = x³ − 3x + 2, both f(1) = 0 and the quotient x² + x − 2 vanish at x = 1, so f(x) = (x − 1)²(x + 2).

What does a factor tell me about the graph?

Each linear factor x − a marks a point where the graph meets the x-axis, at x = a. A single factor makes the curve cross the axis there. A squared factor makes it touch and turn back without crossing.