Splitting the middle term to factorise ax² + bx + c
Mathematics Prep 2 Term 1, Unit 2, Lesson 4
GoalFactorise ax² + bx + c with a ≠ 1 by splitting the middle term into two terms and grouping.
New words
split the middle term · write bx as the sum of two termsproduct a × c · the two numbers must multiply to a × csum b · and add up to bcheck · expand the brackets and compare with the original
Explanation
To factorise 3x² + 8x + 4 without guessing: multiply the first and last coefficients, 3 × 4 = 12. Find two numbers with product 12 and sum 8: they are 2 and 6. Split the middle term: 3x² + 2x + 6x + 4. Now group: x(3x + 2) + 2(3x + 2) = (3x + 2)(x + 2). The table of pairs shows why 2 and 6 are chosen. Signs work as before: if the product a × c is negative, the numbers have different signs. For 2x² + x − 6 the product is −12 and the sum is 1, so the numbers are 4 and −3: 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (x + 2)(2x − 3). If the grouping gives different brackets, try the pair of numbers in the other order.
Factorise 2x² + x − 6. Product −12, sum 1: the numbers are 4 and −3. 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (x + 2)(2x − 3). Check: 2x² − 3x + 4x − 6.
Solve 3x² + 8x + 4 = 0. It factorises as (3x + 2)(x + 2) = 0. So 3x + 2 = 0 or x + 2 = 0, giving x = −2/3 or x = −2. Check x = −2: 12 − 16 + 4 = 0.
Class activity
Number-pair hunt. For 2x² + 7x + 3 list the pairs with product 6 and find the one with sum 7. Split the middle term and factorise. Then do the same for 5x² + 11x + 2.
Practice
1
Factorise 2x² + 7x + 3.
Numbers 1 and 6: 2x² + x + 6x + 3 = (2x + 1)(x + 3).
