GoalReview Unit 2: choose the right method to factorise, check by multiplying, and solve quadratic equations and word problems by factorising.
New words
common factor · always look for the greatest common factor firstdifference of squares · a² − b² = (a − b)(a + b)zero-product rule · if A × B = 0 then A = 0 or B = 0completing the square · add (half the coefficient of x)² to make a perfect square
Explanation
Decide the method by looking at the expression. 1) A common factor in every term: take out the GCF first. 2) Two terms that are both squares with a minus between: a² − b². Two cubes: a³ ± b³ = (a ± b)(a² ∓ ab + b²). 3) Three terms: a perfect square a² ± 2ab + b², or x² + bx + c with two numbers of product c and sum b, or ax² + bx + c by trial or by splitting the middle term. 4) Four terms: group in pairs, then take out the common bracket. 5) If nothing else works, complete the square. To solve an equation, move everything to one side so that the other side is 0, factorise, and use the zero-product rule; check each answer by substitution. Always check a factorisation by multiplying back, and finish the factorisation: x² − 4 inside a bracket must be factorised again. Never divide both sides of x² = 5x by x; you would lose the answer 0.
Solve 2x² − 7x − 15 = 0. Split the middle term: the numbers with product 2 × (−15) = −30 and sum −7 are −10 and 3. So 2x² − 10x + 3x − 15 = 2x(x − 5) + 3(x − 5) = (2x + 3)(x − 5) = 0. Then x = −3/2 or x = 5.
A rectangular garden is x m wide and x + 5 m long with area 84 m². Then x(x + 5) = 84, x² + 5x − 84 = 0 and (x + 12)(x − 7) = 0. A width cannot be negative, so x = 7 m and the length is 12 m. Check: 7 × 12 = 84.
Class activity
Factorisation match. Write six expressions on cards (a common factor, a difference of squares, a trinomial, a perfect square, four terms, a sum of cubes) and their factorised forms on other cards. Match them in pairs and check each by multiplying.
Practice
1
Factorise 3x² − 27.
3(x² − 9) = 3(x − 3)(x + 3).
2
Factorise 27x³ + 1.
(3x + 1)(9x² − 3x + 1).
3
Solve x² + 2x = 35.
x² + 2x − 35 = 0, (x + 7)(x − 5) = 0, so x = −7 or x = 5.
