MathematicsExam code: 0580
Expanding and Factorising Brackets
Algebra and Sequences
- Expand any two brackets first and simplify, then multiply that result by the remaining bracket
- The second stage multiplies a two-term bracket by a three-term expression, so a grid is worth using
- (x + 1)(x + 2)(x + 3) expands to x³ + 6x² + 11x + 6
- The highest power in the answer matches the number of brackets, which is a quick check
Worked example
Three brackets, two at a time
Expand and simplify (3x − 2)(x + 1)(x − 5).
Solution:
- Expand two of the brackets first. Any pair works, so take the two with a plain x: (x + 1)(x − 5)
- x² − 5x + x − 5 = x² − 4x − 5
- Multiply that three-term result by the remaining bracket, (3x − 2): each of its two terms times each of the three, six products in all
- 3x × (x² − 4x − 5) = 3x³ − 12x² − 15x
- −2 × (x² − 4x − 5) = −2x² + 8x + 10
- Collect the like terms: −12x² − 2x² = −14x², and −15x + 8x = −7x
- The answer is 3x³ − 14x² − 7x + 10
- Check with a value: x = 2 gives 4 × 3 × (−3) = −36 in the question, and 24 − 56 − 14 + 10 = −36 in the answer