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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)? P3 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)? P3 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)? P3 1 4a b 1 4a 12 a3
- 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)? P3 1 4a b 1 4a 12 a3 P x x 1 x 3Q 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)? P3 1 4a b 1 4a 12 a3 P x x 1 x 3Q 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 23 2 22 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 23 2 22 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 23 2 22 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) R4 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) R4 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 R4 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 R4 5 ab 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 R4 5 ab 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 R4 5 ab 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 R4 5 ab 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*

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