Week 5 · Factorising: common factor, then trinomials and quadratic equations · 3 lessons
2
Factorising trinomials x² + bx + c
Mathematics Prep 2 Term 1, Unit 2, Lessons 1–2
GoalFactorise x² + bx + c by finding two numbers with product c and sum b, using the sign rules.
New words
trinomial · an expression with three terms, such as x² + 7x + 12product and sum · we need p × q = c and p + q = bsign rules · c > 0: same signs as b; c < 0: different signscheck · expand the two brackets and compare
Explanation
(x + p)(x + q) = x² + (p + q)x + pq. So to factorise x² + bx + c we look for two numbers p and q whose product is c and whose sum is b. For x² + 7x + 12 the pairs with product 12 are 1 × 12, 2 × 6 and 3 × 4, and 3 + 4 = 7, so x² + 7x + 12 = (x + 3)(x + 4). Signs: if c is positive, p and q have the same sign as b (both positive or both negative). If c is negative, p and q have different signs, and the one larger in size takes the sign of b. List the pairs in a table to be sure. If the first coefficient is negative, take out −1 first: −x² + 3x + 40 = −(x² − 3x − 40) = −(x − 8)(x + 5). Always check by expanding.
Pair with product −18
Sum
1, −18
−17
−1, 18
17
2, −9
−7
−2, 9
7
3, −6
−3
−3, 6
3
Worked examples
Factorise x² − 5x + 6. Product 6 and sum −5: both numbers are negative, −2 and −3. So x² − 5x + 6 = (x − 2)(x − 3). Check: x² − 3x − 2x + 6.
Factorise x² − 7x − 18. Product −18, sum −7. The table shows the pair 2 and −9 works. So x² − 7x − 18 = (x + 2)(x − 9).
Factorise x² + 5xy + 6y². The same idea with y: product of numbers 6, sum 5, so the numbers are 2 and 3 and the answer is (x + 2y)(x + 3y). Check: x² + 3xy + 2xy + 6y².
Class activity
Factor-pair cards. Write the pairs of whole numbers with product 24 on cards. For b = 10, 11, 12 and 14 find the pair with sum b and write x² + bx + 24 as two brackets. Which value of b has no pair?
Practice
1
Factorise x² + 9x + 20.
(x + 4)(x + 5).
2
Factorise x² − 8x + 15.
(x − 3)(x − 5).
3
Factorise x² + x − 12.
(x + 4)(x − 3).
4
Factorise x² − 2x − 24.
(x − 6)(x + 4).
5
Karim writes x² + 5x + 6 = (x + 1)(x + 6). Check him by expanding.
(x + 1)(x + 6) = x² + 7x + 6, not x² + 5x + 6. The right pair is 2 and 3: (x + 2)(x + 3).