Products and simple quadratic factors

Extract a common factor first when present

For 6x + 9, both terms contain 3, so 6x + 9 = 3(2x + 3). Expanding checks the factorisation. A common factor must divide every term; a term inside a sum is not itself a factor of the whole expression.

Every term multiplies every term

For (x + 3)(x + 4), distribute one complete bracket: x(x + 4) + 3(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12.

A product grid has four cells: x², 4x, 3x and 12. Omitting either cross product loses part of the original multiplication. For (x-2)(x + 5), keep the signs: x² + 5x-2x-10 = x² + 3x-10.

Reverse a monic quadratic

To factor x² + 7x + 12, find two numbers with sum 7 and product 12: 3 and 4. Thus it equals (x + 3)(x + 4). Expanding checks both the middle term and constant.

For x²-x-12, the required numbers are −4 and 3, so the factors are (x-4)(x + 3). A negative product requires opposite signs. The sum then decides which magnitude has the negative sign.

This lesson factors expressions whose x² coefficient is 1. General quadratic factorisation and shortcut identities need their own additional method; do not infer them merely from a familiar-looking pair of brackets.

Check your understanding 1 Check the middle term

Factor x²-8x + 15. Explain why (x + 3)(x + 5) is wrong.

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The two numbers have product 15 and sum −8, so they are −3 and −5. The expression is (x-3)(x-5). Positive 3 and 5 would produce a middle term of + 8x.

A factorisation is an identity, not yet an equation

If x² + 3x-10 = (x + 5)(x-2), does this alone require x = 2 or x = -5?