Course Hero Logo

4A Chapter 5.pdf - 5 More about Polynomials 5 More about...

Course Hero uses AI to attempt to automatically extract content from documents to surface to you and others so you can study better, e.g., in search results, to enrich docs, and more. This preview shows page 1 out of 47 pages.

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...
View Full Document

Newly uploaded documents

Show More

Newly uploaded documents

Show More

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture