# 11X1 T13 04 polynomial theorems

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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 dividedby (x 2) 11. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (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 dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 1132 13. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 1132 19 14. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 32 19 remainder when P ( x ) is divided by ( x 2) is 19 15. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 32 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 hencefactorise P(x). 17. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 113 2 19 remainder when P ( x ) is divided by ( x 2) is 19 Factor TheoremIf (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 hencefactorise 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 dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 113 2 19 remainder when P ( x ) is divided by ( x 2) is 19 Factor TheoremIf (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 hencefactorise P(x).P 2 2 19 2 30 30 19. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 113 2 19 remainder when P ( x ) is divided by ( x 2) is 19 Factor TheoremIf (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 hencefactorise P(x).P 2 2 19 2 30 30 ( x 2) is a factor 20. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 30P 2 2 19 2 30 30 ( x 2) is a factor 21. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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). x2x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 30 ( x 2) is a factor 22. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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). x2x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3 x3 2 x 20 ( x 2) is a factor 23. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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). x2x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3 x3 2 x 20 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 dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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). x2x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3 x3 2 x 20 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 dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 xx 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 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 dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 xx 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor2 x2 4 x 27. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 xx 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor2 x2 4 x 15 x 28. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 xx 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor2 x2 4 x 15 x 30 29. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 x 15x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor2 x2 4 x 15 x 30 30. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 x 15x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor2 x2 4 x 15 x 3015 x 30 31. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 3 2 19 remainder when P ( x ) is divided by ( x 2) is 19Factor TheoremIf (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 2 x 15x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor2 x2 4 x 15 x 3015 x 30 0 32. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 11 32 19 remainder when P ( x ) is divided by ( x 2) is 19Factor 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 2 x 15x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3x3 2 x 20 2 x 2 19 x 30 ( x 2) is a factor 2 x2 4 x P ( x) ( x 2) x 2 2 x 15 15 x 3015 x 30 0 33. e.g. Find the remainder when P x 5 x 3 17 x 2 x 11 is dividedby (x 2) P x 5 x 3 17 x 2 x 11P 2 5 2 17 2 2 1132 19 remainder when P ( x ) is divided by ( x 2) is 19Factor 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 2x 15 x 2 x 3 0 x 2 19 x 30P 2 2 19 2 30 3 x3 2 x 202x 2 19 x 30 ( x 2) is a factor2 x2 4 x P ( x) ( x 2) x 2 2 x 15 15x 30 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 termconstant constant =leading term=constant 41. OR P x x 3 19 x 30 ( x 2) x 2 leading term leading termconstant constant =leading term=constant 42. OR P x x 3 19 x 30 ( x 2) x 2 leading term leading termconstant constant =leading term=constant 43. OR P x x 3 19 x 30 ( x 2) x 215 leading term leading termconstant constant =leading term=constant 44. ORP x x 3 19 x 30 ( x 2) x 215 leading term leading termconstant constant=leading term=constant If you where to expand out now, how many x would you have? 45. ORP x x 3 19 x 30 ( x 2) x 215 leading term leading termconstant constant=leading term=constant If you where to expand out now, how many x would you have? 15x 46. ORP x x 3 19 x 30 ( x 2) x 215 leading term leading termconstant constant=leading term=constant If you where to expand out now, how many x would you have? 15x How many x do you need? 47. ORP x x 3 19 x 30 ( x 2) x 2 15 leading term leading term constant constant=leading term =constant If you where to expand out now, how many x would you have? 15x How many x do you need? 19x 48. ORP x x 3 19 x 30 ( x 2) x 2 15 leading term leading term constant constant=leading term =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. ORP x x 3 19 x 30 ( x 2) x 2 15 leading term leading term constant constant=leading term =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. ORP x x 3 19 x 30 ( x 2) x 2 15 leading term leading term constant constant=leading term =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?4x4 x 2 ? 51. ORP x x 3 19 x 30 ( x 2) x 2 15 leading term leading term constant constant=leading term =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?4x4 x 2 ? 52. ORP x x 3 19 x 30 ( x 2) x 2 2x 15 leading term leading term constant constant=leading term =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 4 x 2 ? P ( x) ( x 2) x 2 2 x 15 ( x 2)( x 5)( x 3) 53. ORP x x 3 19 x 30 ( x 2) x 2 2x 15 leading term leading term constant constant=leading term =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 4 x 2 ? 54. ORP x x 3 19 x 30 ( x 2) x 2 2x 15 leading term leading term constant constant=leading term =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 4 x 2 ? P ( x) ( x 2) x 2 2 x 15 55. ORP x x 3 19 x 30 ( x 2) x 2 2x 15 leading term leading term constant constant=leading term =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 4 x 2 ? P ( x) ( x 2) x 2 2 x 15 ( x 2)( x 5)( x 3) 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 36Constant factors must be a factor of the constant 58. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 36Constant 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 36Constant 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 36Constant 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 formfactors of the leading coefficient 62. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 63. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 4 4 4 4 4 4 4 4 4 64. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 2 68. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 20 69. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 20 ( x 4) is a factor 70. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 20P( x) 4 x 3 16 x 2 9 x 36 ( x 4) is a factor x 4 71. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 20P( x) 4 x 3 16 x 2 9 x 36 ( x 4) is a factor x 4 4x 2 72. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 20P( x) 4 x 3 16 x 2 9 x 36 ( x 4) is a factor x 4 4x 29 73. (ii) Factorise P x 4 x 3 16 x 2 9 x 36Constant 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 formfactors of the leading coefficient 1 2 3 4 6 9 12 18 36Possibilities = , , , , , , , , 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 tooP 4 4 4 16 4 9 4 36 3 20P( x) 4 x 3 16 x 2 9 x 36 ( x 4) is a factor 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 11b 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 polynomialand a and b are real numbers.When P(x) is divided by (x + 1) the remainder is 11. When P(x) isdivided by (x 3) the remainder is 1.(i) What is the value of b? P 1 11b 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 polynomialand a and b are real numbers.When P(x) is divided by (x + 1) the remainder is 11. When P(x) isdivided by (x 3) the remainder is 1.(i) What is the value of b? P 1 11b 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 polynomialand a and b are real numbers.When P(x) is divided by (x + 1) the remainder is 11. When P(x) isdivided by (x 3) the remainder is 1.(i) What is the value of b? P 1 11b 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 polynomialand a and b are real numbers.When P(x) is divided by (x + 1) the remainder is 11. When P(x) isdivided by (x 3) the remainder is 1.(i) What is the value of b? P 1 11b 11 (ii) What is the remainder when P(x) is divided by (x + 1)(x 3)? P3 1 4a b 1 4a 12a3 81. 2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomialand a and b are real numbers.When P(x) is divided by (x + 1) the remainder is 11. When P(x) isdivided by (x 3) the remainder is 1.(i) What is the value of b? P 1 11b 11 (ii) What is the remainder when P(x) is divided by (x + 1)(x 3)? P3 1 4a b 1 P x x 1 x 3Q x 3 x 8 4a 12a3 82. 2004 Extension 1 HSC Q3b)Let P(x) = (x + 1)(x 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomialand a and b are real numbers.When P(x) is divided by (x + 1) the remainder is 11. When P(x) isdivided by (x 3) the remainder is 1.(i) What is the value of b? P 1 11b 11 (ii) What is the remainder when P(x) is divided by (x + 1)(x 3)? P3 1 4a b 1 P x x 1 x 3Q x 3 x 8 4a 12 R x 3x 8a3 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 3a 19 88. 1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x 4), the quotient isQ(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 isQ(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 isQ(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 isQ(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 isQ(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 isQ(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 isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 54a b 5 97. 1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x 4), the quotient isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 5P 1 54a b 5 98. 1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x 4), the quotient isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 5P 1 54a b 5ab 5 99. 1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x 4), the quotient isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 5P 1 54a b 5ab 5 5a 10a 2 100. 1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x 4), the quotient isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 5P 1 54a b 5ab 5 5a 10a 2 b 3 101. 1994 Extension 1 HSC Q4a)When the polynomial P(x) is divided by (x + 1)(x 4), the quotient isQ(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 theremainder is at most 1, i.e. a linear function.(ii) Given that P(4) = 5 , show that R(4) = 5P(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 5P 1 54a b 5ab 5 5a 10a 2 b 3 R x 2 x 3 102. 2x 1 1 use P 2 Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac, 13, 14, 17, 23*