The factor and remainder theorems

Remainder theorem

When f(x) is divided by (x - a), the remainder is f(a).

When f(x) is divided by (ax - b), the remainder is f(b/a).

Factor theorem

If f(a) = 0, then (x - a) is a factor of f(x). If f(b/a) = 0, then (ax - b) is a factor.

To factorise a cubic, try factors of the constant term (±1, ±2, ...) until f(a) = 0, then divide.

Finding unknowns

Each piece of information gives an equation. Two unknowns need two equations, solved simultaneously.

Worked example

f(x) = 2x³ + px² - 13x + 6 has a factor (x - 2). Find p and factorise f(x).

1. f(2) = 0: 16 + 4p - 26 + 6 = 0, so 4p = 4 and p = 1
2. Divide by (x - 2): 2x³ + x² - 13x + 6 = (x - 2)(2x² + 5x - 3)
3. 2x² + 5x - 3 = (2x - 1)(x + 3)

Answer: p = 1; f(x) = (x - 2)(2x - 1)(x + 3)

Check yourself

1. Find the remainder when x³ - 3x² + 4x - 5 is divided by (x - 2).
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   -1. f(2) = 8 - 12 + 8 - 5 = -1.

2. Find the remainder when 4x³ - 2x + 7 is divided by (2x + 1).
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   7.5. 2x + 1 = 0 when x = -1/2. f(-1/2) = -0.5 + 1 + 7 = 7.5.

3. Which of these is a factor of x³ - 6x² + 11x - 6?
   • (x + 1)
   • (x + 2)
   • (x - 3)
   • (x + 3)
   Show answer

   (x - 3). f(3) = 27 - 54 + 33 - 6 = 0, so (x - 3) is a factor.

4. (x + 2) is a factor of x³ + kx² - 4. Find k.
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   3. f(-2) = -8 + 4k - 4 = 0, so 4k = 12 and k = 3.

5. f(x) = x³ + ax + b. (x - 1) is a factor, and the remainder when f(x) is divided by (x - 2) is 6. Find a.
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   -1. f(1) = 0 gives 1 + a + b = 0. f(2) = 6 gives 8 + 2a + b = 6. Subtracting: 7 + a = 6, so a = -1 (and b = 0).

6. Solve x³ - 2x² - 5x + 6 = 0.
   Show answer

   x = 1, x = -2 or x = 3. f(1) = 0, so (x - 1) is a factor. Dividing gives x² - x - 6 = (x - 3)(x + 2).

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 2 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.