A-level Maths revision · Pure, year 1

The factor theorem

The idea

If f(a)=0f(a) = 0, then (x−a)(x - a) is a factor of f(x)f(x). Once you have one factor, divide (or compare coefficients) to get a quadratic, then factorise that.

  1. Try small values such as 1,−1,2,−21, -1, 2, -2 until f(a)=0f(a) = 0.
  2. Divide f(x)f(x) by (x−a)(x - a) to get a quadratic.
  3. Factorise the quadratic to finish.

Watch out For a factor (2x−1)(2x - 1), substitute x=12x = \frac{1}{2}, not x=1x = 1.

See it worked

Worked example

Fully factorise f(x)=x3−2x2−5x+6f(x) = x^3 - 2x^2 - 5x + 6.

  1. f(1)=1−2−5+6=0f(1) = 1 - 2 - 5 + 6 = 0, so (x−1)(x - 1) is a factor.
  2. Dividing: f(x)=(x−1)(x2−x−6)f(x) = (x - 1)(x^2 - x - 6).
  3. x2−x−6=(x−3)(x+2)x^2 - x - 6 = (x - 3)(x + 2).

Answer (x−1)(x−3)(x+2)(x - 1)(x - 3)(x + 2)

Your turn

Question 1Non-calculator3 marks

(x−2)(x - 2) is a factor of x3+ax2−4x+4x^3 + ax^2 - 4x + 4. Find aa.

Question 2Non-calculator3 marks

Solve x3+2x2−5x−6=0x^3 + 2x^2 - 5x - 6 = 0.

Question 3Non-calculator3 marks

Solve 2x3−3x2−3x+2=02x^3 - 3x^2 - 3x + 2 = 0.

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