11 x1 t15 04 polynomial theorems (2013)

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11 x1 t15 04 polynomial theorems (2013)

  1. 1. Polynomial Theorems
  2. 2. Polynomial Theorems Remainder Theorem If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
  3. 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. 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. 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. 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. 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. 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. 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. 10. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided by (x – 2)
  11. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 34. OR P  x   x 3  19 x  30
  35. 35. OR P  x   x 3  19 x  30  ( x  2)  
  36. 36. OR P  x   x 3  19 x  30  ( x  2)  leading term  leading term =leading term 
  37. 37. OR P  x   x 3  19 x  30  ( x  2)  leading term  leading term =leading term 
  38. 38. OR P  x   x 3  19 x  30  ( x  2)  leading term  leading term =leading term 
  39. 39. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 
  40. 40. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term  constant  constant =constant
  41. 41. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term  constant  constant =constant
  42. 42. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term  constant  constant =constant
  43. 43. OR P  x   x 3  19 x  30  ( x  2)  x 2 leading term  leading term =leading term 15  constant  constant =constant
  44. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 56. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
  57. 57. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36 Constant factors must be a factor of the constant
  58. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 102. 2x 1 1 use P   2 Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac, 13, 14, 17, 23*

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