Grade by Grade
C. Polynomial and Rational Functions · Polynomial Functions and Equations

The Factor Theorem: Factoring Cubics and Quartics

C3.2 C3.1About 75 minutes
Learning goals
  • I can use the factor theorem to decide whether x − b (or ax − b) is a factor of a polynomial.
  • I can list the possible rational zeros of a polynomial with integer coefficients.
  • I can factor cubic and quartic polynomials completely by testing, dividing and factoring the quotient.
  • I can find unknown coefficients so that a given binomial is a factor.

Words to know

factor theorem
x − b is a factor of P(x) if and only if P(b) = 0. More generally, ax − b is a factor if and only if P(b/a) = 0.
integral zero theorem
If x − b is a factor of a polynomial with integer coefficients and leading coefficient 1, then b divides the constant term.
rational zero theorem
If P(x) has integer coefficients and b/a (in lowest terms) is a zero, then b divides the constant term and a divides the leading coefficient.
factor fully
Write as a product of factors that cannot be factored further (here, over the integers).

From remainder to factor

The remainder theorem says the remainder when P(x) is divided by x − b is P(b). If that remainder is 0, then x − b divides P(x) evenly, so it is a factor:

Factor theorem: x − b is a factor of P(x) if and only if P(b) = 0.

"If and only if" means it works both ways: a zero gives a factor, and a factor gives a zero. That matches what you saw on graphs: each x-intercept b goes with a factor (x − b).

Similarly, ax − b is a factor if and only if P(b/a) = 0.

xy−3−124−8−448−2: x + 21: x − 13: x − 3P(x) = x³ − 2x² − 5x + 6
Each zero of P(x) = x³ − 2x² − 5x + 6 gives a factor, so P(x) = (x + 2)(x − 1)(x − 3).

Which values should you test?

You could test forever, so narrow the list.

  • Integral zero theorem: if the leading coefficient is 1 and all coefficients are integers, any integer zero must divide the constant term. For x³ − 2x² − 5x + 6, test only ±1, ±2, ±3, ±6.
  • Rational zero theorem: in general, a rational zero b/a must have b dividing the constant term and a dividing the leading coefficient. For 2x³ + 3x² − 8x + 3, the candidates are ±1, ±3, ±1/2, ±3/2.

Start with easy values such as 1 and −1. A handy check: P(1) is just the sum of the coefficients.

1. List candidates± (factors of constant) ÷ (factors of leading coeff.)2. Test with P(b)substitute until P(b) = 03. Divide by (x − b)synthetic division; remainder must be 04. Factor the quotienttrinomial factoring, or repeat steps 1-35. Checkexpand, or test a value of x
A strategy for factoring a cubic or quartic with the factor theorem.
Worked example

Example 1: Testing a factor

Is x − 2 a factor of P(x) = x³ − 3x² + 4? If so, factor P(x) fully.

  1. P(2) = 8 − 12 + 4 = 0, so yes, x − 2 is a factor.
  2. Synthetic division with 2 on 1, −3, 0, 4: 1, −1, −2 | 0. Quotient x² − x − 2.
  3. x² − x − 2 = (x − 2)(x + 1).

P(x) = (x − 2)²(x + 1)

Worked example

Example 2: Factoring a cubic

Factor x³ − 2x² − 5x + 6 fully.

  1. Candidates: ±1, ±2, ±3, ±6.
  2. P(1) = 1 − 2 − 5 + 6 = 0 ✓ (the coefficients add to 0), so x − 1 is a factor.
  3. Divide by x − 1: 1, −1, −6 | 0. Quotient x² − x − 6 = (x − 3)(x + 2).

x³ − 2x² − 5x + 6 = (x − 1)(x − 3)(x + 2)

Worked example

Example 3: Factoring a quartic

Factor x⁴ − 5x³ + 5x² + 5x − 6 fully.

  1. P(1) = 1 − 5 + 5 + 5 − 6 = 0, so x − 1 is a factor. Divide: 1, −4, 1, 6 | 0, giving x³ − 4x² + x + 6.
  2. Now factor the cubic. Test −1: −1 − 4 − 1 + 6 = 0, so x + 1 is a factor. Divide by x + 1: 1, −5, 6 | 0, giving x² − 5x + 6.
  3. x² − 5x + 6 = (x − 2)(x − 3).

x⁴ − 5x³ + 5x² + 5x − 6 = (x − 1)(x + 1)(x − 2)(x − 3)

Worked example

Example 4: Leading coefficient not 1

Factor 2x³ + 3x² − 8x + 3 fully.

  1. Candidates: ±1, ±3, ±1/2, ±3/2.
  2. P(1) = 2 + 3 − 8 + 3 = 0, so x − 1 is a factor. Divide: 2, 5, −3 | 0, giving 2x² + 5x − 3.
  3. 2x² + 5x − 3 = (2x − 1)(x + 3). (Check: 2x² + 6x − x − 3 ✓.)

2x³ + 3x² − 8x + 3 = (x − 1)(2x − 1)(x + 3). (The zero ½ was on the candidate list.)

Worked example

Example 5: Making a factor

Find k so that x + 2 is a factor of x³ + kx² − x − 2, then factor fully.

  1. Factor theorem: P(−2) = 0. P(−2) = −8 + 4k + 2 − 2 = 4k − 8 = 0, so k = 2.
  2. P(x) = x³ + 2x² − x − 2. Group: x²(x + 2) − 1(x + 2) = (x + 2)(x² − 1).

k = 2 and P(x) = (x + 2)(x − 1)(x + 1)

Worked example

Example 6: When the quotient will not factor

Factor x³ − 3x² − 2x + 4 as far as possible over the integers.

  1. P(1) = 1 − 3 − 2 + 4 = 0, so x − 1 is a factor. Divide: 1, −2, −4 | 0, giving x² − 2x − 4.
  2. Discriminant: (−2)² − 4(1)(−4) = 20. It is not a perfect square, so x² − 2x − 4 has no integer factors. Its zeros are 1 ± √5 (irrational).

x³ − 3x² − 2x + 4 = (x − 1)(x² − 2x − 4), fully factored over the integers.

Just for fun

The factor theorem in one sentence: when a polynomial hits zero, it hands you a factor. In math, getting a zero is sometimes the best result you can get.

Hands-on activity

Factor detective cards

You need: 12 small cards or scraps of paper, a pencil and a calculator.

Case file: P(x) = 2x³ − 3x² − 11x + 6.

  1. Write every candidate zero b/a on its own card: b divides 6 and a divides 2.
  2. Test the cards one at a time with P(b). Put cards with P(b) ≠ 0 in a "cleared" pile.
  3. As soon as you find a zero, divide and factor the quotient. Which cards did you not even need to test?
  4. Write P(x) fully factored and check by expanding.

What to look for: candidates ±1, ±2, ±3, ±6, ±1/2, ±3/2 (12 cards). P(3) = 54 − 27 − 33 + 6 = 0, so x − 3 is a factor; dividing gives 2x² + 3x − 2 = (2x − 1)(x + 2). So P(x) = (x − 3)(2x − 1)(x + 2), with zeros 3, ½ and −2. After one success, the quotient finishes the job, so most cards never need testing.

Your turn

1

Is x + 1 a factor of x³ + 2x² − x − 2?

2

Is x − 3 a factor of 2x³ − 5x² − 4x + 3?

3

Factor x³ + 2x² − 5x − 6 fully.

4

Factor x³ − 7x + 6 fully.

5

List the possible rational zeros of 3x³ − 4x² + x + 2.

6

Factor 2x³ − 5x² − 4x + 3 fully.

7

Find k so that x − 2 is a factor of x³ − kx² + 3x + 2.

More practice: warm-up

1

P(4) = 0. Name a factor of P(x).

2

List the possible integer zeros of x³ + 3x² − 2x − 10.

3

Is x − 1 a factor of x³ − 1?

4

P(−2) = 5. Is x + 2 a factor of P(x)?

5

3x − 2 is a factor of P(x) exactly when P(?) = 0.

More practice: on your own

1

Factor x⁴ − 2x³ − 13x² + 14x + 24 fully.

2

Factor 3x³ + x² − 8x + 4 fully.

3

Factor x³ − 4x² + x + 2 as far as possible over the integers.

4

Find a and b so that x − 1 and x + 2 are both factors of x³ + ax² + bx − 6. Then factor fully.

5

Without calculating P(2), explain why x − 2 cannot be a factor of x³ + 4x² + 3x + 7.

Challenge

1

Factor 6x³ + 7x² − x − 2 fully.

2

Use the factor theorem to show that x − a is a factor of xⁿ − aⁿ for every whole number n ≥ 1.

3

Factor x⁴ + x³ − 7x² − x + 6 fully.

Check your understanding

Quiz

1.

Which is a factor of P(x) = x³ − 4x² + x + 6?

2.

Which number cannot be a rational zero of 2x³ + x² − 5x + 3?

3.

What is the fully factored form of x³ − x² − 4x + 4?

4.

For which k is x + 1 a factor of x³ + kx + 3?

5.

P(x) has integer coefficients and leading coefficient 1, and its constant term is 15. Which could be an integer zero?

6.

Synthetic division of P(x) by x − 4 gives a remainder of 0 and quotient x² + 2x − 3. What is P(x) factored fully?

For parents and teachers

Big idea (C3.2): the factor theorem says x − b is a factor exactly when P(b) = 0. Factoring a cubic or quartic becomes a search: list the possible rational zeros, test them, divide out the factor you find, and factor what is left.

How to help: Ask your teen to show you their list of candidates before they start testing, and to check the final answer by expanding or by substituting a value (for example x = 2) into both the original and the factored form.

Common difficulties:

  • Testing x = 2 and then writing the factor as (x + 2).
  • Leaving out fractional candidates when the leading coefficient is not 1.
  • Forgetting to keep factoring the quotient.
  • Arithmetic errors in P(b) with negative values; a calculator helps.