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# 11 x1 t15 04 polynomial theorems (2013)

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• 1. Polynomial Theorems
• 2. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
• 3. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x)
• 4. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x) let A( x)  ( x  a )
• 5. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x) let A( x)  ( x  a ) P  x   ( x  a )Q( x)  R( x)
• 6. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x) let A( x)  ( x  a ) P  x   ( x  a )Q( x)  R( x) P  a   (a  a )Q(a )  R (a )
• 7. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x) let A( x)  ( x  a ) P  x   ( x  a )Q( x)  R( x) P  a   (a  a )Q(a )  R (a )  R(a)
• 8. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x) let A( x)  ( x  a ) P  x   ( x  a )Q( x)  R( x) P  a   (a  a )Q(a )  R (a )  R(a) now degree R ( x)  1  R ( x) is a constant
• 9. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a) Proof: P  x   A( x)Q( x)  R( x) let A( x)  ( x  a ) P  x   ( x  a )Q( x)  R ( x) P  a   (a  a )Q(a )  R (a )  R(a) now degree R ( x)  1  R ( x) is a constant R( x)  R(a)  P(a)
• 10. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2)
• 11. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11
• 12. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11 3 2
• 13. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19 3 2
• 14. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2
• 15. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0
• 16. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence factorise P(x).
• 17. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence factorise P(x). P  2    2   19  2   30 3
• 18. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence factorise P(x). P  2    2   19  2   30 0 3
• 19. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence factorise P(x). P  2    2   19  2   30 0  ( x  2) is a factor 3
• 20. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 0  ( x  2) is a factor
• 21. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x2 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 0  ( x  2) is a factor
• 22. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x2 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0  ( x  2) is a factor
• 23. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x2 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2x 2  ( x  2) is a factor
• 24. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x2 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2x 2 19 x  30  ( x  2) is a factor
• 25. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor
• 26. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x
• 27. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15 x
• 28. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15 x 30
• 29. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x 15 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15 x 30
• 30. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x 15 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15 x 30 15 x  30
• 31. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x 15 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15 x 30 15 x  30 0
• 32. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2 x 15 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2 x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15 x 30  P ( x)  ( x  2)  x 2  2 x  15  15 x  30 0
• 33. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2) P  x   5 x 3  17 x 2  x  11 P  2   5  2   17  2   2  11  19  remainder when P ( x ) is divided by ( x  2) is  19 3 2 Factor Theorem If (x – a) is a factor of P(x) then P(a) = 0 e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence x 2 2x 15 factorise P(x). x  2 x 3  0 x 2  19 x  30 3 P  2    2   19  2   30 x3  2 x 2 0 2x 2 19 x  30  ( x  2) is a factor 2 x2  4 x 15x 30  P ( x)  ( x  2)  x 2  2 x  15  15 x  30  ( x  2)( x  5)( x  3) 0
• 34. OR P  x   x 3  19 x  30
• 35. OR P  x   x 3  19 x  30  ( x  2)  
• 36. OR P  x   x 3  19 x  30  ( x  2)  leading term  leading term =leading term 
• 37. OR P  x   x 3  19 x  30  ( x  2)  leading term  leading term =leading term 
• 38. OR P  x   x 3  19 x  30  ( x  2)  leading term  leading term =leading term 
• 39. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 
• 40. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term  constant  constant =constant
• 41. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term  constant  constant =constant
• 42. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term  constant  constant =constant
• 43. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant
• 44. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant If you where to expand out now, how many x would you have?
• 45. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant If you where to expand out now, how many x would you have? 15x
• 46. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant If you where to expand out now, how many x would you have? 15x How many x do you need?
• 47. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant If you where to expand out now, how many x would you have? 15x How many x do you need? 19x
• 48. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need?
• 49. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4x
• 50. OR P  x   x 3  19 x  30  ( x  2)  x 2 15  constant  constant =constant leading term  leading term =leading term If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4 x  2  ? 4x
• 51. OR P  x   x 3  19 x  30  ( x  2)  x 2 15  constant  constant =constant leading term  leading term =leading term If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4 x  2  ? 4x
• 52. OR P  x   x 3  19 x  30  ( x  2)  x 2 2x 15  constant  constant =constant leading term  leading term =leading term If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4 x  2  ?  P ( x)  ( x  2)  x 2  2 x  15   ( x  2)( x  5)( x  3) 4x
• 53. OR P  x   x 3  19 x  30  ( x  2)  x 2 2x 15  constant  constant =constant leading term  leading term =leading term If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4 x  2  ? 4x
• 54. OR P  x   x 3  19 x  30  ( x  2)  x 2 2x 15  constant  constant =constant leading term  leading term =leading term If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4 x  2  ?  P ( x)  ( x  2)  x 2  2 x  15  4x
• 55. OR P  x   x 3  19 x  30  ( x  2)  x 2 2x 15  constant  constant =constant leading term  leading term =leading term If you where to expand out now, how many x would you have? 15x How many x do you need? 19x How do you get from what you have to what you need? 4 x  2  ?  P ( x)  ( x  2)  x 2  2 x  15   ( x  2)( x  5)( x  3) 4x
• 56. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
• 57. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant
• 58. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
• 59. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!!
• 60. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! Fractional factors must be of the form
• 61. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient
• 62. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4
• 63. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4
• 64. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2
• 65. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2
• 66. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too
• 67. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 3 2
• 68. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 0 3 2
• 69. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 0  ( x  4) is a factor 3 2
• 70. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 0  ( x  4) is a factor 3 2 P( x)  4 x 3  16 x 2  9 x  36   x  4  
• 71. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 0  ( x  4) is a factor 3 2 P( x)  4 x 3  16 x 2  9 x  36   x  4  4x 2 
• 72. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 0  ( x  4) is a factor 3 2 P( x)  4 x 3  16 x 2  9 x  36   x  4  4x 2 9 
• 73. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 of course they could be negative!!! factors of the constant Fractional factors must be of the form factors of the leading coefficient 1 2 3 4 6 9 12 18 36 Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 1 2 3 4 6 9 12 18 36 = , , , , , , , , 2 2 2 2 2 2 2 2 2 they could be negative too P  4   4  4   16  4   9  4   36 0  ( x  4) is a factor 3 2 P( x)  4 x 3  16 x 2  9 x  36   x  4  4x 2 9    x  4  2 x  3 2 x  3
• 74. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b?
• 75. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11
• 76. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11
• 77. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11 (ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?
• 78. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11 (ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)? P3  1
• 79. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11 (ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)? P3  1 4a  b  1
• 80. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11 (ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)? P3  1 4a  b  1 4a  12 a3
• 81. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11 (ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)? P3  1 4a  b  1 4a  12 a3 P x    x  1 x  3Q x   3 x  8
• 82. 2004 Extension 1 HSC Q3b) Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial and a and b are real numbers. When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is divided by (x – 3) the remainder is 1. (i) What is the value of b? P 1  11 b  11 (ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)? P3  1 4a  b  1 4a  12 a3 P x    x  1 x  3Q x   3 x  8  R x   3x  8
• 83. 2002 Extension 1 HSC Q2c) Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial. Find the value of a.
• 84. 2002 Extension 1 HSC Q2c) Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial. Find the value of a. P  2   3
• 85. 2002 Extension 1 HSC Q2c) Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial. Find the value of a. P  2   3  23  2 22  a  3
• 86. 2002 Extension 1 HSC Q2c) Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial. Find the value of a. P  2   3  23  2 22  a  3  16  a  3
• 87. 2002 Extension 1 HSC Q2c) Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial. Find the value of a. P  2   3  23  2 22  a  3  16  a  3 a  19
• 88. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b?
• 89. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function.
• 90. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5
• 91. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x)
• 92. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
• 93. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5
• 94. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
• 95. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) R4  5
• 96. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) R4  5 4a  b  5
• 97. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) P 1  5 R4  5 4a  b  5
• 98. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) P 1  5 R4  5 ab 5 4a  b  5
• 99. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) P 1  5 R4  5 ab 5 4a  b  5  5a  10 a  2
• 100. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) P 1  5 R4  5 ab 5 4a  b  5  5a  10 a  2 b  3
• 101. 1994 Extension 1 HSC Q4a) When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is Q(x) and the remainder is R(x). (i) Why is the most general form of R(x) given by R(x) = ax + b? The degree of the divisor is 2, therefore the degree of the remainder is at most 1, i.e. a linear function. (ii) Given that P(4) = – 5 , show that R(4) = – 5 P(x) = (x + 1)(x – 4)Q(x) + R(x) P(4) = (4 + 1)(4 – 4)Q(4) + R(4) R(4) = – 5 (iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x) P 1  5 R4  5 ab 5 4a  b  5  5a  10 R x   2 x  3 a  2 b  3
• 102. 2x 1 1 use P   2 Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac, 13, 14, 17, 23*