Let us begin with x² + 7x + 12 = x² + (a + b)x + ab.

Step-by-step solution

Idea: This is splitting the middle term without tiles. Since (x + a)(x + b) = x² + (a + b)x + ab, we need two numbers whose sum is the coefficient of x (7) and whose product is the constant (12).

1. Compare x² + 7x + 12 with x² + (a + b)x + ab: the x-coefficients give a + b = 7 and the constants give ab = 12.½ mark
2. Pairs of whole numbers with product 12: 1 × 12 (sum 13), 2 × 6 (sum 8), 3 × 4 (sum 7) ✓.½ mark
3. So a = 3, b = 4 (or a = 4, b = 3, which gives the same factors). The middle term splits as 7x = 3x + 4x.½ mark
4. x² + 7x + 12 = (x + 3)(x + 4)½ mark

Check: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Choosing 2 and 6 because 2 × 6 = 12, without checking that their sum is 7 (it is 8).
• Choosing numbers that add to 7 but do not multiply to 12, such as 1 and 6, or 2 and 5.
• Writing the answer as (x + 3) + (x + 4). Factors are multiplied, not added.

How this can come in the exam

Try one yourself

Factorise x² + 8x + 12.

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