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Unformatted text preview: 5 More about Polynomials 5 More about Polynomials (3) (1) : (6 x 4 y ) (6 x 7 y ) 20 (8) 3 y 12 Review Exercise 5 (p. 5.5) y 4 By substituting y = –4 into (2), we have 3x 2(4) 10 1. (a) (b) f (0) (0) 2 4(0) 9 009 9 f (2) (2) 2 4(2) 9 6. ∴ 3x 10 8 x 6 The solution is x = –6, y = –4. (a) 3ab 6a 2b 4 8 9 3ab(1 2a) 5 (b) (c) f (3) (3) 2 4(3) 9 9 12 9 a(3 y 1) x 3xy a(3 y 1) x(3 y 1) (3 y 1)(a x) 30 2. (a) 7. f (1) 6(1) 3(1) 1 (a) 3 x 2 ( x 4) 2( x 4) 6 3 1 ( x 4)( x 2 2) 4 (b) x3 4x 2 2x 8 (b) f (2) 6(2) 3 3(2) 1 4a 10bc 5c 8ab 4a 8ab 5c 10bc 48 6 1 (4a 8ab) (5c 10bc) 41 4a(1 2b) 5c(1 2b) (1 2b)(4a 5c) 3 (c) 3. 1 1 1 f 6 3 1 2 2 2 6 3 1 8 2 1 4 ∵ h(3) 14 ∴ (3) 2 4(3) a 14 8. 9. (a) x 2 4 x 12 ( x 2)( x 6) (b) 4a 2 7ab 15b 2 (4a 5b)(a 3b) (a) 8x 2 2 2(4 x 2 1) 9 12 a 14 2(2 x 1)(2 x 1) a 7 4. (b) 2 x y 5 ......(1) x 3 y 10 ......(2) From (1), we have y 5 2 x ......(3) By substituting (3) into (2), we have x 3(5 2 x) 10 x 15 6 x 10 4[(2a) 2 2(2a)(5b) (5b) 2 ] 4(2a 5b) 2 10. (a) 1 64x 3 1 ( 4 x )3 5 x 5 (1 4 x)[12 (1)(4 x) (4 x) 2 ] x 1 By substituting x = 1 into (3), we have y 5 2(1) 3 ∴ The solution is x = 1, y = 3. 5. 16a 2 80ab 100b 2 4(4a 2 20ab 25b 2 ) (1 4 x)(1 4 x 16x 2 ) (b) 16a 3 54b 3 2(8a 3 27b3 ) 6 x 7 y 8 ......(1) 2[(2a)3 (3b)3 ] 3x 2 y 10 ......(2) (2) 2 : 6x 4 y 20......( 3) 2(2a 3b)[(2a) 2 (2a)(3b) (3b) 2 ] 2(2a 3b)(4a 2 6ab 9b 2 ) 149 NSS Mathematics in Action (2nd edition) 4A Full Solutions 11. 49a 2 7a 28ab Activity (7 a ) 2 7 a (1 4b) 7a 1 4b Activity 5.1 (p. 5.19) 1. Remainder (by long division) (a) 2 2x2 ( y 1) 2 3 3( y 1) 3 6x x 9( y 1) (b) 0 (c) 6 x x x 1 2 y y y x( x 1) y ( y 1) y x 1 x( y 1) (d) 10 2x ( y 1) 6x 3( y 1) 3 2 12. 2 2 13. 2. f (1) 3(1) 2 (1) 4 2 f (1) 3(1) 2 (1) 4 0 f (2) 3(2) 2 (2) 4 6 f (2) 3(2) 2 (2) 4 10 Yes, the remainder of f ( x) ( x a) is equal to f (a). Activity 5.2 (p. 5.30) 1. (a) ma (b) nc 5 x 1 4( x 1) 5 x 1 4x 4 5 x 1 4x 1 x 1 14. 4 15. Value of f(a) 2. (a) 1, 3 (b) 6x 2 18x 3 x x 3 6x2 18x x3 x 3 6 x 2 18x x3 6 x( x 3) x3 6x 1, 2 3. x 1, x 2, x 1, x 2, 3x 1, 3x 2, 3x 1, 3x 2 4. x 1 or x 2 or 3x 1 5. ( x 1)( x 2)(3x 1) Maths Dialogue Maths Dialogue (p. 5.14) 1. (a) When the degree of the remainder is less than that of the divisor, the division process should stop. x (b) 1 2 2x 1 2x2 2x 1 4x 6x 16. 2 x 3 3x 2 4 x(3 x 2) 6 x(2 x 3) (2 x 3)(3 x 2) 2x2 x x 1 x 12x 2 8 x 12x 2 18x (2 x 3)(3 x 2) 10x (2 x 3)(3 x 2) ∴ 150 1 2 1 2 1 1 Quotient x , remainder 2 2 5 More about Polynomials 1 5 x 2 4 2. 2x2 x 3 (b) x 2 x3 x 2 3x 2 x 1 x 2 3x 2 1 x2 x 2 2 x3 x 2 3x x2 5 x 2 2 5 5 x 2 4 3x 3x Remainder 0 3 4 x 5 3 ∴ Quotient , remainder 2 4 4 (c) 3x 2 4 x 3 2x 6x 8x 2 6x 1 3 6x3 Maths Dialogue (p. 5.34) 1. The factor theorem is not applicable when the polynomial has no linear factor(s). 2. (a) Quotient 2 x 2 x 3 ∴ 8x 2 6x 1 8x 2 6x 1 4 x8 4 x 4 1 6x (2 x ) 2(2 x ) 1 4 2 4 1 (2 x 4 1) 2 ∴ (b) Quotient 3x 4 x 3 Remainder 1 x 5x 9 4 2 2 ( x 4 6 x 2 9) x 2 Classwork (p. 5.15) (a) By division algorithm, we have f ( x) (2 x 3)(2 x 3) 4 ( x 2 3) 2 x 2 ( x 2 3 x)( x 2 3 x) (2 x 3) 2 4 ( x 2 x 3)( x 2 x 3) 4 x 2 12x 9 4 4 x 2 12x 13 Classwork Classwork (p. 5.7) 1 1. 2 x 2 , 3 x 3 xy, 0.6 2 2. x4 4 5 0 0 (a) (b) (c) (d) Coefficient of x3 x2 –6 3 0 –2 2 0 0 1 x –1 1 4 –2 (b) By division algorithm, we have f ( x) (3x 4)(5 x 1) (2) (3x 4)(5 x) (3x 4)(1) 2 Constant term 5 –7 –5 0 15x 2 20x 3x 4 2 Degree of polynomial 4 4 3 2 15x 2 23x 2 Classwork (p. 5.20) (a) f(3) (b) f(4) Classwork (p. 5.11) (a) 3x 2 x 3x 2 2 x (c) f(–1) (d) f(–7) 3x 2 Classwork (p. 5.21) 1 (a) f 3 2x 2x ∴ Quotient 3x 2 Remainder 0 151 (b) 2 f 5 (c) 1 f 4 NSS Mathematics in Action (2nd edition) 4A Full Solutions 5 f 2 Alternative Solution 4d 2 3d 1 ) d 3 Classwork (p. 5.38) (a) H.C.F. a 2 4d 3 3d 2 d (d) ) L.C.M. a 5 (b) 4 d 3 9 d 2 8d 3 a 5b a 5 b Quick Practice 5.4 (p. 5.13) 2x2 x 2 a 3b 2 a 3 b 2 H.C.F. a 3 b ∴ x 3 2 x3 5 x 2 5 x 2 a b 3 2 x3 6 x 2 L.C.M. a b 5 a b 5 (c) 12d 2 9d 3 2 x2 5x 2 2 x 2 3x 2x 2 ab2c 2 a b 2 c 2 2x 6 a3c a3 c ∴ H.C.F. a c ac 8 ∴ Quotient 2 x x 2 , remainder 8 2 L.C.M. a b c 3 2 2 Quick Practice 5.5 (p. 5.13) 2x2 5x 5 a 3b 2 c 2 2 x 2 4 x3 6 x 2 0 x 3 Quick Practice 4 x3 4 x 2 10x 2 0 x 3 Quick Practice 5.1 (p. 5.7) (3t 2 7t ) (2t 2 2t 9) 10x 2 10x 3t 7t 2t 2t 9 2 2 10x 3 3t 2 2t 2 7t 2t 9 10x 10 t 2 5t 9 7 Alternative Solution 3t 2 7t 0 ∴ Quotient 2 x 5x 5 , remainder 7 2 ) 2t 2 2t 9 Quick Practice 5.6 (p. 5.13) 5x 3 2 3 2 x x 1 5x 2x 4x 3 t 2 5t 9 5 x3 5 x 2 5 x Quick Practice 5.2 (p. 5.8) (3x 3 8 x 2 x) ( x 2 2 x 3 5 x 2) 3x 2 9 x 3 3x 8 x x x 2 x 5 x 2 3 2 2 3 3x 2 3x 3 3x 2 x 8 x x x 5 x 2 3 3 2 2 12x 6 ∴ Quotient 5 x 3 , remainder 12x 6 x3 7 x 2 6x 2 Alternative Solution 3x3 8 x 2 x 0 Quick Practice 5.7 (p. 5.14) x2 3x 2 0 x 2 3x3 6 x 2 x 1 ) 2 x x 5 x 2 3 2 x3 7 x 2 6 x 2 3x3 0 x 2 2 x 6x2 x 1 Quick Practice 5.3 (p. 5.9) (1 3d 4d 2 )(d 3) 6x2 0x 4 (1 3d 4d )(d ) (1 3d 4d )(3) 2 x3 ∴ Quotient x 2 , remainder x 3 2 d 3d 2 4d 3 3 9d 12d 2 4d 3 3d 2 12d 2 d 9d 3 Quick Practice 5.8 (p. 5.16) Let p(x) be the required polynomial. By division algorithm, we have 4d 3 9d 2 8d 3 152 5 More about Polynomials 3 x3 8 x 2 6 p( x) (3 x 2 2 x 4) 2 Quick Practice 5.12 (p. 5.23) Let f ( x) 25x 3 sx2 10x 6. By the remainder theorem, (3 x 8 x 6) 2 3x 2 2 x 4 3 3x 8 x 2 8 2 3x 2 x 4 ( x 2)(3x 2 2 x 4) 3x 2 2 x 4 x2 ∴ The required polynomial is x + 2. p( x) 3 2 1 f 4 5 3 Quick Practice 5.9 (p. 5.16) By division algorithm, we have ax3 5 x 2 8 x (2 x 2 x 3)(3x 1) (bx 3) (2 x 2 x 3)(3x) (2 x 2 x 3)(1) bx 3 6 x3 3x 2 9 x 2 x 2 x 3 bx 3 6 x3 5 x 2 (b 10) x By comparing the coefficient of x 3 on both sides, a6 By comparing the coefficient of x on both sides, 8 b 10 b2 Quick Practice 5.10 (p. 5.20) Let f ( x) x 3 x 2 6 x 5. (a) By the remainder theorem, remainder f (2) Quick Practice 5.13 (p. 5.23) Let f ( x) 6 x 2 4 x 1. By the remainder theorem, 3 f (b) 2 3 2 6b 4b 1 2 1 2 6b 4b 0 2 12b 2 8b 1 0 (6b 1)(2b 1) 0 b (2) 3 (2) 2 6(2) 5 8 4 12 5 3 1 1 or b 6 2 Quick Practice 5.14 (p. 5.24) Let f ( x) x 3 cx 2 dx 8. When f(x) is divided by x – 2, remainder 14 f (2) 14 (b) By the remainder theorem, remainder f (4) (2)3 c(2) 2 d (2) 8 14 (4) 3 (4) 2 6(4) 5 64 16 24 5 51 8 4c 2d 8 14 2c d 7 ......(1) When f(x) is divided by x + 1, remainder 17 f (1) 17 Quick Practice 5.11 (p. 5.22) Let f ( x) 9 x 3 3x 2 4. (a) By the remainder theorem, 1 remainder f 3 (1) 3 c(1) 2 d (1) 8 17 1 c d 8 17 c d 8 ......(2) 3 (1) + (2): 3c 15 c 5 2 1 1 9 3 4 3 3 1 1 4 3 3 4 By substituting c = –5 into (2), we have 5 d 8 d 3 (b) By the remainder theorem, 2 remainder f 3 3 2 1 1 1 25 s 10 6 4 5 5 5 1 s 264 5 25 s 1 25 5 s 5 Quick Practice 5.15 (p. 5.27) (a) f (2) 2(2)3 9(2) 2 12(2) 4 16 36 24 4 0 ∴ x – 2 is a factor of f(x). 2 2 2 9 3 4 3 3 8 4 4 3 3 0 153 NSS Mathematics in Action (2nd edition) 4A Full Solutions (b) (b) ∵ f (3) 0 ∴ By the factor theorem, x + 3 is a factor of f(x). By long division, 2x2 x 1 3 x 3 2x 7 x2 2x 3 f (1) 2(1)3 9(1) 2 12(1) 4 2 9 12 4 27 0 ∴ x + 1 is not a factor of f(x). 2 x3 6 x 2 3 (c) 2 1 1 1 1 f 2 9 12 4 2 2 2 2 1 9 64 4 4 0 ∴ 2x – 1 is a factor of f(x). x2 2x 3 x 2 3x x3 x3 ∴ ( x 3)( x 1)(2 x 1) Quick Practice 5.16 (p. 5.28) Let f ( x) 2 x 3 x 2 kx 3. ∵ f(x) is divisible by 2x – 1, i.e. 2x – 1 is a factor of f(x). 1 f 0 ∴ 2 3 f ( x) ( x 3)(2 x 2 x 1) Quick Practice 5.19 (p. 5.31) ∵ f (1) (1)3 6(1)2 11(1) 6 0 ∴ x – 1 is a factor of f(x). By long division, x2 5 x 6 2 1 1 1 2 k 3 0 2 2 2 1 1 k 30 4 4 2 k 7 2 2 k 7 x 1 x 3 6 x 2 11x 6 x3 x 2 5 x 2 11x 6 5x2 5x 6x 6 6x 6 Quick Practice 5.17 (p. 5.29) ∵ f(x) is divisible by 2x – 3, i.e. 2x – 3 is a factor of f(x). 3 ∴ f 0 2 ∴ f ( x) ( x 1)( x 5 x 6) 2 ( x 1)( x 3)( x 2) Quick Practice 5.20 (p. 5.32) ∵ f (1) 2(1) 3 13(1) 2 13(1) 10 12 0 2 3 3 2 p q 0 2 2 9 3 pq0 2 2 9 3 p 2q 0 f (1) 2(1) 3 13(1) 2 13(1) 10 18 0 f (2) 2(2)3 13(2) 2 13(2) 10 0 ∴ x – 2 is a factor of f(x). By long division, 2x2 9x 5 3 p 2q 9 ......(1) When f(x) is divided by x + 2, remainder 21 f (2) 21 x 2 2 x 3 13x 2 13x 10 2 x3 4 x 2 9 x 2 13x 10 2(2) p(2) q 21 2 9 x 2 18x 8 2 p q 21 5 x 10 2 p q 29 ......(2) (1) 2 (2) : 7 p 49 5 x 10 ∴ p7 f ( x) ( x 2)(2 x 2 9 x 5) ( x 2)( x 5)(2 x 1) By substituting p = 7 into (2), we have 2(7) q 29 Quick Practice 5.21 (p. 5.33) (a) 4 x 3 6 x 2 28x 30 2(2 x 3 3x 2 14x 15) q 15 Let f ( x) 2 x 3 3x 2 14x 15. Quick Practice 5.18 (p. 5.29) 3 2 (a) f (3) 2(3) 7(3) 2(3) 3 ∵ ∴ 54 63 6 3 0 154 f (1) 2(1)3 3(1) 2 14(1) 15 0 x – 1 is a factor of f(x). 5 More about Polynomials By long division, 2x2 Quick Practice 5.24 (p. 5.40) (a) 6 x 2 4 x 2 2(3x 2 2 x 1) x 15 2(3x 1)( x 1) x 1 2 x 3 3 x 2 14x 15 2x 2x 3 (1 3x) (3x 3) 3(3x 1) 2 ( x 1) ∴ H.C.F. (3x 1)( x 1) 2 2 x 2 14x 15 x2 L.C.M. 6(3x 1)2 ( x 1) x 15x 15 15x 15 ∴ x 3 x x( x 2 1) x( x 1)( x 1) (b) 4 x 6 x 28x 30 2(2 x 3x 14x 15) 3 2 3 2 x 2 3x 4 ( x 1)( x 4) 2( x 1)(2 x 2 x 15) 2( x 1)( x 3)(2 x 5) x 8 x 2 16x x( x 2 8 x 16) 3 x( x 4) 2 (b) 4 x3 6 x 2 28x 30 0 ∴ 2( x 1)( x 3)(2 x 5) 0 (by (a)) x 1 0 or x 3 0 or 2 x 5 0 x 1 or x 3 or x H.C.F. 1 L.C.M. x( x 1)( x 1)( x 4) 2 5 2 Quick Practice 5.25 (p. 5.40) (a) (i) f (3) (3)3 4(3) 2 4(3) 3 27 36 12 3 Quick Practice 5.22 (p. 5.38) (a) 4a3b 2c 22 a3 b 2 c 6a 4 c 2 2 3 a 4 ∴ 0 (ii) ∵ f (3) 0 (by (a)) ∴ x – 3 is a factor of f(x). By long division, f ( x) ( x 3)( x 2 x 1) c2 H.C.F. 2 a 3 c 2a 3 c L.C.M. 2 2 3 a 4 b 2 c 2 12a 4 b 2 c 2 (b) f ( x) ( x 3)( x2 x 1) g ( x) x 3 1 (b) ( x 5)(5 x 1) ( x 5) (5 x 1) 3 (5 x 1) ( x 5) 2 ∴ ( x 1)( x 2 x 1) (5 x 1) ( x 5) 2 H.C.F. (5 x 1) 3 2 2 ∴ 2 H.C.F. x 2 x 1 L.C.M. ( x 3)( x 1)( x 2 x 1) L.C.M. ( x 5) (5 x 1) 3 ( x 5) 2 ( x 5)(5 x 1) 3 ( x 5) 2 Quick Practice 5.26 (p. 5.44) 16b 1 6b 9b 2 (a) 2 2 6b 1 2b 3b Quick Practice 5.23 (p. 5.39) 5(2 x 1)( x 1) 5 (2 x 1) ( x 1) (a) (1 b)(1 3b) 8b 1 b 10(2 x 1)3 2 5 (2 x 1)3 25( x 1) (2 x 1) 3 ∴ 5 (2 x 1) ( x 1) 2 3 H.C.F. 5(2 x 1) L.C.M. 2 52 (2 x 1) 3 ( x 1) ( x 1) 3 (b) 50(2 x 1) 3 ( x 1)( x 1) 3 (b) 4(2 x y ) 2 (3x 2 y ) 22 (2 x y ) 2 (3x 2 y ) 6(3x 2 y ) 2 2 3 8(2 x y )3 (3x 2 y ) 23 ∴ 8 16b (3x 2 y ) 2 (2 x y )3 (3x 2 y ) H.C.F. 2(3x 2 y) 3 3 2 L.C.M. 2 3 (2 x y ) (3x 2 y ) 24(2 x y )3 (3x 2 y ) 2 155 (1 3b) 2 2(1 3b) x2 4 x 2 12x x 2 7 x 12 2 x 2 8 x2 4 x( x 3) ( x 3)( x 4) 2( x 2 4) 2 x2 4 x( x 3) ( x 3)( x 4) 2( x 2)( x 2) 2x ( x 4)( x 2) NSS Mathematics in Action (2nd edition) 4A Full Solutions Quick Practice 5.27 (p. 5.44) x 1 3x 2 2 x 1 (a) 2 3x 1 3x x x 1 (3 x 1)( x 1) x(3 x 1) 3x 1 x 1 3x 1 x(3 x 1) (3 x 1)( x 1) 1 x(3 x 1) (b) (b) 3r 9 r 3 3r 2 2 2 r 7 r 3 2 r 5r 3 3(r 3) r 2 (r 3) (2r 1)(r 3) (2r 1)(r 3) 3(r 3) (2r 1)(r 3) (2r 1)(r 3) r 2 (r 3) 3 2 r 2 Quick Practice 5.30 (p. 5.46) 4x 2 1 4 x 2 25 y 2 2 x 5 y 2 x 5 y Quick Practice 5.28 (p. 5.45) 1 6 (a) x 4 x 2 6x 8 1 6 x 4 ( x 2)( x 4) x2 6 ( x 2)( x 4) ( x 2)( x 4) x4 ( x 2)( x 4) (b) 4x 6 x2 4 x 2 9 3x 2 5 x 2 2(2 x 3) x2 (2 x 3)(2 x 3) ( x 2)(3 x 1) 2 1 2 x 3 3x 1 2(3 x 1) 2x 3 (2 x 3)(3 x 1) (2 x 3)(3 x 1) 2(3 x 1) (2 x 3) (2 x 3)(3 x 1) 6x 2 2x 3 (2 x 3)(3 x 1) 4x 1 (2 x 3)(3 x 1) 8 1 25 16x 2 4 x 2 5 x 8 1 (5 4 x)(5 4 x) x(5 4 x) 8x 5 4x x(5 4 x)(5 4 x) x(5 4 x)(5 4 x) 5 4x x(5 4 x)(5 4 x) 1 x(5 4 x) 4x 2 1 (2 x 5 y )(2 x 5 y ) 2 x 5 y 2 x 5 y 4x 2(2 x 5 y ) 2x 5 y (2 x 5 y )(2 x 5 y ) (2 x 5 y )(2 x 5 y ) (2 x 5 y )(2 x 5 y ) 4 x 2(2 x 5 y ) (2 x 5 y ) (2 x 5 y )(2 x 5 y ) 4 x 4 x 10 y 2 x 5 y (2 x 5 y )(2 x 5 y ) 6 x 15 y (2 x 5 y )(2 x 5 y ) 3(2 x 5 y ) (2 x 5 y )(2 x 5 y ) 3 2x 5 y Quick Practice 5.31 (p. 5.47) A B R.H.S. x 3 3x 5 A(3 x 5) B( x 3) ( x 3)(3 x 5) (3 A B) x (3B 5 A) ( x 3)(3x 5) By comparing the like terms in the numerators on both sides, we have 2 3A B ......(1) Quick Practice 5.29 (p. 5.46) 6 2x 1 (a) 6x 3 2x 2 7 x 4 2 6 2x 1 3(2 x 1) (2 x 1)( x 4) 2 1 2x 1 x 4 2( x 4) 2x 1 (2 x 1)( x 4) (2 x 1)( x 4) 2( x 4) (2 x 1) (2 x 1)( x 4) 2x 8 2x 1 (2 x 1)( x 4) 9 (2 x 1)( x 4) 2 5 A 3B ......(2) 3 (1) (2) : 4 4 A A 1 By substituting A = 1 into (1), we have 2 3(1) B B 1 156 5 More about Polynomials 2. Quick Practice 5.32 (p. 5.47) 1 4x 2 x4 x2 2x x2 4 2x2 x 1 2(2 x 1) x4 x( x 2) ( x 2)( x 2) x(2 x 1) 1 2( x 4) x( x 2) x( x 2)( x 2) x2 2x 8 x( x 2)( x 2) x( x 2)( x 2) 3x 6 x( x 2)( x 2) 3( x 2) x( x 2)( x 2) 3 x ( x 2) (a) By division algorithm, we have 4 x3 8 x 2 a ( x 2 2)(4 x b) (8 x 9) ( x 2 2)(4 x) ( x 2 2)(b) 8 x 9 4 x 3 8 x bx2 2b 8 x 9 4 x 3 bx2 2b 9 By comparing the coefficient of x 2 on both sides, b 8 By comparing the constant term on both sides, a 2(8) 9 a7 2x 2 (b) 2x 4 4x 3 8x 2 7 4x 3 8x 2 7 Further Practice ∴ Quotient 2x 2 , remainder 7 Further Practice (p. 5.9) 1. ( x 2 2 x 1)(1 x) 3 Further Practice (p. 5.24) 1. Let f ( x) x 99 2 x 1. By the remainder theorem, remainder f (1) ( x 2 2 x 1)(1) ( x 2 2 x 1)( x) 3 x 2 2 x 1 x3 2 x 2 x 3 (1) 99 2(1) 1 1 2 1 2 x3 x 2 2 x 2 2 x x 1 3 x3 x 2 3x 2 2. (3x 3 4 x 5 x 2 ) ( x 2 1)(2 x 1) 2. 3x 3 4 x 5 x 2 [( x 2 1)(2 x) ( x 2 1)(1)] 3x 3 4 x 5 x 2 (2 x 3 2 x x 2 1) 3x3 4 x 5 x 2 2 x3 2 x x 2 1 3( a) 2 6(a ) 1 1 a 3x 2 x x x 4 x 2 x 5 1 3 3 2 Let f ( x) 3x 2 6 x 1. By the remainder theorem, f (a) 1 a 2 3a 2 6a 1 1 a x3 2 x 4 3a 2 5a 2 0 (a 2)(3a 1) 0 Further Practice (p. 5.17) 1. (a) 3x 2 3x 2 0 x 1 9 x 3 6 x 2 0 x 4 a 2 or a 9 x 3 0 x 2 3x 3. 6 x 2 3x 4 6x 2 0x 2 3 ∴ Quotient 3x 2 , remainder 3x 2 (b) 3x 2 (2 3x) 4 6x 2 9x3 4 9x3 6x 2 4 3x 2 3x 2 9 x 3 6 x 2 0 x 4 9x 6x By the remainder theorem, 1 1 f g 2 2 2 3 2 1 1 1 1 1 1 k 4 2 7 4 2 k 3 2 2 2 2 2 2 k 1 1 k 1 1 7 3 8 2 2 2 5 k 5 8 k 8 3x 2 3 1 3 Further Practice (p. 5.30) 1. Let f ( x) x 2 2ax 4 . ∵ f(x) is divisible by x + a, i.e. x + a is a factor of f(x). 2 4 ∴ Quotient 3x , remainder 4 2 157 NSS Mathematics in Action (2nd edition) 4A Full Solutions f (a) 0 ∴ (b) L.H.S. f ( x) g ( x) 2( x 1)( x 6) ( x 1)( x 2)(2 x 3) ( a ) 2a ( a ) 4 0 2 2( x 1) 2 ( x 2)( x 6)(2 x 3) R.H.S. H.C.F. L.C.M. a 2a 4 0 2 2 a2 4 0 ( x 1) 2( x 1)( x 2)( x 6)(2 x 3) (a 2)(a 2) 0 2( x 1) 2 ( x 2)( x 6)(2 x 3) ∵ L.H.S. = R.H.S. ∴ f ( x) g ( x) H.C.F. L.C.M. a 2 or a 2 2. ∵ ∴ x – 2 is a common factor of f(x) and g(x). f (2) 0 and g (2) 0 f (2) 0 Further Practice (p. 5.48) 5x 3 y xy y 1. 5 x 2 8 x 3 x 2 2 xy x 2 y 5x 3 y y ( x 1) ( x 1)(5 x 3) x( x 2 y ) x 2 y x( x 2 y ) y 1 y x 2y x (2) 2 p(2) q 0 4 2p q 0 2 p q 4 ......(1) g (2) 0 (2) 3 p(2) q 0 82p q 0 2 p q 8 ......(2) (1) (2) : 4 p 4 p 1 2. By substituting p = 1 into (1), we have 2(1) q 4 q 6 Further Practice (p. 5.40) 1. (a) 4 x 2 12x 9 (2 x 3) 2 9 x 4 32 x 4 ∴ H.C.F. 1 L.C.M. 33 x 4 (2 x 3) 3 27x 4 (2 x 3) 3 2 x 2 xy y 2 ( x y )(2 x y ) 4 x 4 xy y (2 x y) 2 2 3. 2 ( x 2 y 2 )(2 x y ) ( x y )( x y )(2 x y ) ∴ H.C.F. 2 x y L.C.M. ( x y)( x y)(2 x y) 2 (a) 3 2( x 3) 10 x 3 ( x 2)( x 3) ( x 2)( x 3) 3x 1 x 2 x 10 2 x 5 x 10x 25 x 1 ( x 1) 2 2 x5 x 1 2( x 1) x5 f ( x) 4 x 2 14x 12 2( x 2 7 x 6) 2( x 1)( x 6) ∵ x 3 x 1 2( x 5) 2 x 1 x 5 ( x 5) x( x 5) 3 x 1 2( x 5) 2 x 1 ( x 5 ) ( x 5) 2 2 x 2 x 1 2( x 5) x 1 ( x 5) 2 2 x3 x 2 y 2 xy 2 y 3 x 2 (2 x y ) y 2 (2 x y ) 2. 3( x 3) 2 10 ( x 3) 2 x 2 ( x 2)( x 3) 3 2x 4 x 3 ( x 2)( x 3) 3 2( x 2) x 3 ( x 2)( x 3) 3 2 x3 x3 5 x3 3 33 (2 x 3)3 (b) (6 x 9) [3(2 x 3)] 3 3x 9 10 2 x 2 6x 9 2 x x 2 x 6 g (1) 2(1)3 9(1) 2 13(1) 6 30 0 Exercise g (1) 2(1)3 9(1) 2 13(1) 6 0 ∴ x + 1 is a factor of g(x). By long division, g ( x) ( x 1)(2 x 2 7 x 6) Exercise 5A (p. 5.9) Level 1 1. (3x 2 4 x 12) (2 x 2 4 x 12) ( x 1)( x 2)(2 x 3) ∴ H.C.F. x 1 3x 2 4 x 12 2 x 2 4 x 12 3x 2 2 x 2 4 x 4 x 12 12 L.C.M. 2( x 1)( x 2)( x 6)(2 x 3) 5 x 2 24 158 5 More about Polynomials 2. 12. ( x 2 8)(1 4 x 2 x 2 ) ( x 3 2 x 2 5x 18) (4 x 3 12x 2 9 x 5) x 3 2 x 2 5 x 18 4 x 3 12x 2 9 x 5 ( x 2 8)(1) ( x 2 8)(4 x) ( x 2 8)(2 x 2 ) x 3 4 x 3 2 x 2 12x 2 5 x 9 x 18 5 x 2 8 4 x 3 32x 2 x 4 16x 2 5 x 10x 4 x 13 8 32x x 2 16x 2 4 x 3 2 x 4 3 2 8 32x 15x 2 4 x 3 2 x 4 3. (4 x x 8) (3x x 2) 2 2 4 x 2 x 8 3x 2 x 2 Level 2 13. (1 x 3x 2 ) (1 3x x 2 ) (3 x 2 ) 4 x 3x x x 8 2 2 4. 2 x 2 2x 6 1 x 3x 2 1 3x x 2 3 x 2 ( x 3 x 2 3x 10) (2 x 3 5x 2 2 x 6) 1 4 x 5x 2 1 1 3 x 3x 3x 2 x 2 x 2 x 3 x 2 3x 10 2 x 3 5 x 2 2 x 6 14. (5 5x 2 x 3 ) (4 x 2 6 x 1) (2 5x 4 x 3 ) x 3 2 x 3 x 2 5 x 2 3x 2 x 10 6 5. x 3 4 x 2 5x 4 5 5x 2 x 3 4 x 2 6x 1 2 5x 4 x 3 (5x 2 10 3x) (14x 2 x 3 2 x 2 8) 6 x 9 x 2 5x 3 5 1 2 6 x 5x 5x 2 4 x 2 x 3 4 x 3 5 x 2 10 3x 14x 2 x 3 2 x 2 8 15. (2 x 3 x 2 x 1) (3x 5x 3 4) (3x 3 5x 2 ) 2 x 3 5 x 2 2 x 2 3x 14x 10 8 6. 2 x 3 7 x 2 11x 2 2 x 3 x 2 x 1 3x 5 x 3 4 3x 3 5 x 2 (3x 3 7 x 6 x 2 9) (18 5x 3 7 x 6 x 2 ) 7 x 6x 2 7 x3 1 4 3 x 3x 3x x 2 5 x 2 2 x 3 5 x 3 3x 3 7 x 6 x 2 9 18 5 x 3 7 x 6 x 2 16. ( x 2)( x 4)(3x 4) [( x 2)( x) ( x 2)(4)](3 x 4) 3x 3 5 x 3 6 x 2 6 x 2 7 x 7 x 9 18 2x3 9 7. ( x 2 2 x 4 x 8)(3 x 4) ( x 2 2 x 8)(3x 4) (2 x 8x 2 5) ( x 3 6 x 2 20 2 x) ( x 2 2 x 8)(3x) ( x 2 2 x 8)(4) 2 x 8 x 2 5 x 3 6 x 2 20 2 x 3 x 3 6 x 2 24x 4 x 2 8 x 32 x 3 8 x 2 6 x 2 2 x 2 x 5 20 3 x 3 6 x 2 4 x 2 24x 8 x 32 x 3 14x 2 4 x 15 8. 3 x 3 10x 2 16x 32 (25 x 3 5x 2 3x) (12x 5x 2 3x 3 7) 17. (2 x 1)(1 2 x)(2 2 x) [(2 x 1)(1) (2 x 1)(2 x)](2 2 x) 25 x 3 5 x 2 3x 12x 5 x 2 3x 3 7 x 3 3x 3 5 x 2 5 x 2 3x 12x 25 7 (2 x 1 4 x 2 2 x)(2 2 x) 4 x 3 15x 18 9. (4 x 2 4 x 1)(2 2 x) (4 x 2 4 x 1)(2) (4 x 2 4 x 1)(2 x) (2 3x)(6 x 2 ) 8 x 2 8 x 2 8 x 3 8 x 2 2 x (2 3x)(6) (2 3x)( x 2 ) 8x 3 8x 2 8x 2 8x 2 x 2 12 18x 2 x 2 3x 3 8 x 3 16x 2 10x 2 10. (5 2 x)(6 2 x x 2 ) 18. ( x 2)(2 x 5) ( x 2 3) (5 2 x)(6) (5 2 x)(2 x) (5 2 x)( x 2 ) [( x 2)(2 x) ( x 2)(5)] ( x 2 3) 30 12x 10x 4 x 2 5 x 2 2 x 3 (2 x 2 4 x 5 x 10) x 2 3 30 2 x x 2 2 x 3 2 x 2 x 10 x 2 3 2 x 2 x 2 x 10 3 11. (2 x x 2 3)(3 x) x2 x 7 (2 x x 2 3)(3) (2 x x 2 3)( x) 6 x 3x 2 9 2 x 2 x 3 3x 9 6 x 3x 3x 2 2 x 2 x 3 9 9x x 2 x3 159 NSS Mathematics in Action (2nd edition) 4A Full Solutions (b) ∵ Coefficient of x2 = 5 a 35 ∴ a 8 19. (3x 2) 2 3x(1 2 x) (9 x 2 12x 4) 3x 6 x 2 9 x 2 6 x 2 12x 3x 4 ∵ Constant term = –12 6 b 12 ∴ b 18 3x 2 15x 4 20. ( x 3)( x 3) 2( x 3 x 2 3x 3) ( x 2 9) 2( x3 x 2 3x 3) x 2 9 2 x3 2 x 2 6 x 6 (2 x a)(bx 1) 2( x 3) [(2 x a)(bx) (2 x a)(1)] 2( x 3) 2 x3 x 2 2 x 2 6 x 9 6 (2bx2 abx 2 x a) 2 x 6 2 x3 3x 2 6 x 3 2bx2 abx 2 x a 2 x 6 26. (a) 2bx2 abx 2 x 2 x a 6 21. (2 x 1)( x 3x 1) (1 x ) 2 2bx2 (ab 4) x (6 a) 2 [(2 x 1)( x 2 ) (2 x 1)(3x) (2 x 1)(1)] (1 x 2 ) (b) ∵ Coefficient of x2 = 6 ∴ 2b 6 (2 x 3 x 2 6 x 2 3x 2 x 1) 1 x 2 (2 x 3 7 x 2 x 1) 1 x 2 b3 ∵ Constant term = –2 ∴ 6 a 2 a 8 ∴ The coefficient of x (8)(3) 4 2x3 7 x 2 x 1 1 x 2 2x3 7 x 2 x 2 x 1 1 2 x 3 8x 2 x 2 20 22. (5 x)(2 x 2 3x 1) x(5x 4) [(5 x)(2 x ) (5 x)(3x) (5 x)(1)] x(5 x 4) 2 Exercise 5B (p. 5.17) Level 1 x2 1. 2x 2x2 4x 3 (10x 2 2 x 3 15x 3x 2 5 x) 5 x 2 4 x 10x 2 2 x 3 15x 3x 2 5 x 5 x 2 4 x 2 x 3 10x 2 3x 2 5 x 2 15x x 4 x 5 2 x 3 8 x 2 12x 5 2x2 4x 3 23. (3x 2 1)(3x 1) (1 x)(2 x 2 x 1) 4x [(3x 2 1)(3x) (3x 2 1)(1)] 3 [(1 x)(2 x 2 ) (1 x)( x) (1 x)(1)] ∴ Quotient x 2 , remainder 3 (9 x 3 3x 3x 2 1) (2 x 2 2 x 3 x x 2 1 x) 9 x 3 3x 3x 2 1 2 x 2 2 x 3 x x 2 1 x 3 2 2 x 3 4 x 8 x 4 x 12x 7 9 x 2 x 3x 2 x x 3x x x 1 1 3 2x2 2. 3 2 7 x 3 5x 2 2 8x3 4 x 2 12x 7 24. (2 x 2 x 1)( x 1) 2 x 2 (3 x) 4x2 [(2 x x 1)( x) (2 x x 1)(1)] 2 x (3 x) 2 2 2 12x 7 12x (2 x 3 x 2 x 2 x 2 x 1) 6 x 2 2 x 3 2x3 x 2 x 2x 2 x 1 6x 2 2x3 7 2x3 2x3 x 2 2x 2 6x 2 x x 1 ∴ 3x 2 2 x 1 ∵ The degree of the polynomial is 2. ∴ The claim is disagreed. Quotient 2 x x 3 , remainder 7 2 2x 2 3. 25. (a) ( x ax) ( x 6) (3x b) 2 7 x 2x 0x 7x 2 3 2 2 2x3 x ax x 6 3x b 7x 2 x 2 x 2 ax 3x 6 b 7x 2 2 2 x (a 3) x (6 b) 2 2 ∴ 160 Quotient 2 x 7 , remainder 2 2 5 More about Polynomials 6x 7 4. ∴ Quotient 2 x 1 , remainder 2 x 2 6 x 5 x 14 2 6 x 2 12x 2x 3 4x 6x 6x 1 3 7 x 14 7 x 14 ∴ 3 2x2 10. 2 4 x3 6 x 2 6x 1 Quotient 6 x 7 , remainder 0 6x 9 8 x 4x 1 2 5. ∴ x 3 x x 11x 6 3 2 x 3x 3 Quotient 2 x 3 , remainder 8 2 2 11. 4 x 2 11x 6 x 2 3x 2 3x 1 3x 8 x 2 9 x 1 3 4 x 2 12x 3x3 x 2 x6 9x2 9x 1 x3 9 x 2 3x 3 ∴ 6x 1 Quotient x 2 4 x 1 , remainder 3 3 1 3x 2 6. 6x 2 ∴ x 4 3 x 3 12x 2 x 3 3 x 12x 3 2 2 x 2 3x 4 12. x3 2x 3 4x3 0x 2 x4 x 5 4x3 6x 2 1 ∴ Quotient x 3x 2 , remainder 3 2 6x2 x 5 Quotient 3x 1 , remainder 1 2 6x2 9x 2x2 x 1 8x 5 x 4 2 x 9 x 5 x 15 8 x 12 7. 3 2 17 2 x3 8 x 2 ∴ x 5 x 15 2 Quotient...
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