What are the Answers to all theses questions?

Example 12 Self Tutor The use of brackets is Expand and simplify: essential! a 1 - 2(x + 3)2 6 2 ( 3 + 20) - (2 + 2) (3 - 20 ) a 1 - 2(x +3)2 b 2(3 + 2) - (2 + 2) (3 - 20) =1 - 2x2 + 6x +9] - 6+2x - [6 - 2x + 3x -x2] =1 - 2x2 - 12x - 18 = 6+ 2x - 6+ 2x - 3x + 22 = -2x2 - 12x -17 = 2 + x 5 Expand and simplify: a 1+ 2(x + 3)2 b 2+3(x - 2) (2 + 3) c 3 - (3 - 2)2 d 5 - (2+ 5) (2 - 4) e 1+2(4-2)2 f x2 - 3x - (2 + 2) (2 - 2) g (a + 2)2 - (2+ 1) (2 -4) h (2xx + 3)2+ 3(2 + 1)2 i x2 +3x -2(x - 4)2 i (3x - 2)2 - 2(2 + 1) 2EXERCISE H 1 Fully factorise: a 3x2 + 9x 6 2x2 + 7x 4x2 - 10x d 6x2 - 15x e 9x2 - 25 f 16x2 - 1 9 2x2 - 8 h 3x2 - 9 i 4x2 - 20 j x2 - 8x + 16 k x2 - 10x + 25 1 2x2 - 8x +8 m 16x2 + 40x + 25 n 9x2 + 12x + 4 o x2 - 22x + 121 Example 14 Self Tutor Fully factorise: a 3x2 + 12x + 9 6 -202 + 3x+ 10 a 3x2 + 12x +9 {3 is a common factor} = 3(202 + 42 + 3) { sum = 4, product = 3} = 3(2+ 1) (2+3) b - 202+ 3x+10 = -[22 - 3x - 10] {removing -1 as a common factor to make the coefficient of x2 be 1} = -(2-5)(2+2) {sum = -3, product = -10} 2 Fully factorise: a x2 + 9x + 8 b x2 + 7x+12 G x2 - 7x- 18 d x2+ 4x -21 e x2 - 9x + 18 f x2 + x-6 9 - 202 + 2+ 2 h 3x2 - 42x + 99 i -2x2 - 4x - 2 BACKGROUND KNOWLEDGE 13 j 2x2+ 6x - 20 k 2x2 - 10x - 48 1 -2x2 + 14x - 12 m -3x2 + 6x - 3 n -x2 - 2x - 1 o -5x2 + 10x + 403 Fully factorise: a 2x2 + 5x - 12 b 3x2 - 5x - 2 c 7x2 - 9x + 2 d 6x2 -x -2 e 4x2 -4x - 3 f 10x2 - x - 3 g 2x2 -11x - 6 h 3x2 -5x - 28 i 8x2 + 2x - 3 j 10x2 - 9x - 9 k 3x2 + 23x - 8 1 6x2 + 7x + 2 m -4x2 - 2x + 6 n 12x2 - 16x - 3 o -6x2 - 9x + 42 p 21x - 10 - 9x2 9 8x2 -6x - 27 r 12x2 + 13x + 3 s 12x2 + 20x + 3 t 15x2 - 22x + 8 u 14x2 - 11x - 15 Example 16 Self Tutor Fully factorise: 3(x + 2) + 2(x - 1)(x + 2) - (x + 2)2 3(x + 2) + 2(2 - 1)(2+2) - (2+2)2 = (2+ 2)[3+2(2-1) - (x+2)] {as (x + 2) is a common factor} = (2 + 2) [3+ 22 - 2 - 20-2] = (2+ 2) (20-1) 4 Fully factorise: a 3(ac + 4) + 2(2 + 4) (2 - 1) 6 8(2 - 2) - 3(x + 1)(2 -2) c 6(2 + 2)2 + 9(20 + 2) d 4(x + 5) + 8(2+ 5)2 e (2 + 2) (2 + 3) - (2+ 3) (2-20) f (a + 3)2 + 2(2 + 3) - 20(20 + 3) 9 5(x - 2) - 3(2 - 2) (2 + 7) h 3(1 - x) +2(x +1)(x -1) Example 17 Self Tutor Fully factorise using the 'difference of two squares': a (2 + 2) 2 - 9 6 (1 - 20) 2 - ( 220 + 1) 2 a (20 + 2) 2 - 9 b (1 - 20) 2 - (22 + 1)2 = (2+ 2)2-32 = [(1 - ac) + (2x + 1)][(1 - 2) -(22 + 1)] = [(2+ 2) + 3][(2+2) -3] = [1 - a + 2x + 1][1 - x - 2x - 1] = (2+5)(2-1) =-32(2+ 2)