AS Mathematics Algebra Manipulation of brackets Objectives Be confident in the use of brackets Be able to factorise linear expressions Review of expanding brackets Example 1 Expand (x + 3)(x + 5) x2 + 5x + 3x + 15
x2 + 8x + 15 multiply term by term collect like terms Alternatives for expanding (x + 3)(x + 5) smiley face (x+3)(x+5) grid method x
Difference of 2 squares! x2 - 64 Remember (a-b)(a+b) = a2 - b2 Example 4 A harder one! Expand
(x2 + 2x + 1)(x - 2) x3 - 2x2 + 2x2 4x + x - 2 x3 -3x -2 multiply term by term collect like terms Example 5 Expand
(x + 4)(x - 3)(2x + 1) (x2 + x 12)(2x + 1) 2x3 + x2 + 2x2 + x 24x - 12 2x3 + 3x2 23x - 12 multiply any two brackets multiply remaining bracket collect like terms Factorising This involves taking out any common factors.
Try to spot the HCF by inspection. (i) common factors Example 1 Factorise 12x 18y 6(2x 3y) Example 2 Factorise 6x2 21x 3x(2x 7) The HCF of 12x & 18y is 6
Check your answer by expanding the brackets The HCF of 6x2 & 21x is 3x (ii) grouping Example 3 Factorise ax + ay + bx + by a(x + y) + b(x + y) The first two terms have common factor a, the last two
have common factor y There is now a common factor of (x + y) (x + y)(a + b) Check your answer by expanding the brackets. To illustrate this:a(x + y) + b(x + y) let z = x + y az + bz z(a + b) but z = x + y (x + y)(a + b) ..as before!
Example 4 Factorise xy + 2x + 3y + 6 x(y + 2) + 3(y + 2) The first two terms have common factor x, the last two have common factor 3 There is now a common factor of (y + 2) (y + 2)(x + 3) Check your answer by expanding the brackets.
Example 5 Factorise 2x - 2xy - y + y2 2x(1 - y) - y(1 - y) (1 - y)(2x - y) For this method to succeed, both brackets should be the same, i.e both (1 - y)
Example 6 Factorise 6a + 3ab - 2b - b2 Check your answer by 3a(2 + b) - b(2 + b) expanding the brackets (2 + b)(3a - b)
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