Expanding and Factorising Brackets

Algebra and Sequences

For x² + bx + c, look for a pair of numbers that multiply to give c and add to give b
Those two numbers go straight into the brackets
- x² + 7x + 12 needs numbers multiplying to 12 and adding to 7, which are 3 and 4, giving (x + 3)(x + 4)
- x² − 3x − 10 needs numbers multiplying to −10 and adding to −3, which are −5 and 2, giving (x − 5)(x + 2)
A negative constant means the two numbers have opposite signs

Worked example

Factorising a simple quadratic

Factorise x² + 2x − 15.

Solution:

Factorising reverses expanding, so the answer is a pair of brackets
With no number in front of x², the two brackets each start with x, and the two numbers in them must multiply to −15 and add to +2
Those two conditions come straight from the expansion: the numbers multiply to give the constant and add to give the x coefficient
The factor pairs of 15 are 1 × 15 and 3 × 5
The product needed is negative, so the two numbers have opposite signs
Test the pairs: 5 and −3 give 5 × (−3) = −15 and 5 + (−3) = 2, which is what is wanted
So x² + 2x − 15 = (x + 5)(x − 3)
Check by expanding back: x² − 3x + 5x − 15 = x² + 2x − 15
The order of the brackets does not matter, but the signs inside them do