The document discusses two polynomial theorems: 1) The Remainder Theorem states that if a polynomial P(x) is divided by (x - a), the remainder is P(a). This is proven through algebraic manipulation. 2) The Factor Theorem states that if (x - a) is a factor of a polynomial P(x), then P(a) = 0. An example demonstrates finding a factor of a polynomial and factorizing it.

1. Polynomial Theorems

2. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)

3. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)

4. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)
let A( x)  ( x  a )

5. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)
let A( x)  ( x  a )
P  x   ( x  a )Q( x)  R( x)

6. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)
let A( x)  ( x  a )
P  x   ( x  a )Q( x)  R( x)
P  a   (a  a )Q(a )  R (a )

7. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)
let A( x)  ( x  a )
P  x   ( x  a )Q( x)  R( x)
P  a   (a  a )Q(a )  R (a )
 R(a)

8. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)
let A( x)  ( x  a )
P  x   ( x  a )Q( x)  R( x)
P  a   (a  a )Q(a )  R (a )
 R(a)
now degree R ( x)  1
 R ( x) is a constant

9. Polynomial Theorems
Remainder Theorem
If the polynomial P(x) is divided by (x – a), then the remainder is P(a)
Proof:
P  x   A( x)Q( x)  R( x)
let A( x)  ( x  a )
P  x   ( x  a )Q( x)  R ( x)
P  a   (a  a )Q(a )  R (a )
 R(a)
now degree R ( x)  1
 R ( x) is a constant
R( x)  R(a)
 P(a)

10. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)

11. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11

12. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
3
2

13. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
3
2

14. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2

15. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0

16. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
factorise P(x).

17. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
factorise P(x).
P  2    2   19  2   30
3

18. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
factorise P(x).
P  2    2   19  2   30
0
3

19. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
factorise P(x).
P  2    2   19  2   30
0
 ( x  2) is a factor
3

20. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
0
 ( x  2) is a factor

21. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x2
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
0
 ( x  2) is a factor

22. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x2
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
 ( x  2) is a factor

23. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x2
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2x 2
 ( x  2) is a factor

24. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x2
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2x 2 19 x  30
 ( x  2) is a factor

25. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor

26. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x

27. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15 x

28. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15 x 30

29. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x 15
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15 x 30

30. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x 15
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15 x 30
15 x  30

31. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x 15
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15 x 30
15 x  30
0

32. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2 x 15
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2 x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15 x 30
 P ( x)  ( x  2)  x 2  2 x  15 
15 x  30
0

33. e.g. Find the remainder when P  x   5 x 3  17 x 2  x  11 is divided
by (x – 2)
P  x   5 x 3  17 x 2  x  11
P  2   5  2   17  2   2  11
 19
 remainder when P ( x ) is divided by ( x  2) is  19
3
2
Factor Theorem
If (x – a) is a factor of P(x) then P(a) = 0
e.g. (i) Show that (x – 2) is a factor of P  x   x 3  19 x  30 and hence
x 2 2x 15
factorise P(x).
x  2 x 3  0 x 2  19 x  30
3
P  2    2   19  2   30
x3  2 x 2
0
2x 2 19 x  30
 ( x  2) is a factor
2 x2  4 x
15x 30
 P ( x)  ( x  2)  x 2  2 x  15 
15 x  30
 ( x  2)( x  5)( x  3)
0

34. OR
P  x   x 3  19 x  30

35. OR
P  x   x 3  19 x  30
 ( x  2) 


36. OR
P  x   x 3  19 x  30
 ( x  2) 
leading term  leading term
=leading term


37. OR
P  x   x 3  19 x  30
 ( x  2) 
leading term  leading term
=leading term


38. OR
P  x   x 3  19 x  30
 ( x  2) 
leading term  leading term
=leading term


39. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term


40. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term

constant  constant
=constant

41. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term

constant  constant
=constant

42. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term

constant  constant
=constant

43. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant

44. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant
If you where to expand out now, how many x would you have?

45. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant
If you where to expand out now, how many x would you have? 15x

46. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant
If you where to expand out now, how many x would you have? 15x
How many x do you need?

47. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x

48. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?

49. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
leading term  leading term
=leading term
15 
constant  constant
=constant
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4x

50. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
15 
constant  constant
=constant
leading term  leading term
=leading term
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4 x  2  ?
4x

51. OR
P  x   x 3  19 x  30
 ( x  2)  x 2
15 
constant  constant
=constant
leading term  leading term
=leading term
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4 x  2  ?
4x

52. OR
P  x   x 3  19 x  30
 ( x  2)  x 2 2x 15 
constant  constant
=constant
leading term  leading term
=leading term
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4 x  2  ?
 P ( x)  ( x  2)  x 2  2 x  15 
 ( x  2)( x  5)( x  3)
4x

53. OR
P  x   x 3  19 x  30
 ( x  2)  x 2 2x 15 
constant  constant
=constant
leading term  leading term
=leading term
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4 x  2  ?
4x

54. OR
P  x   x 3  19 x  30
 ( x  2)  x 2 2x 15 
constant  constant
=constant
leading term  leading term
=leading term
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4 x  2  ?
 P ( x)  ( x  2)  x 2  2 x  15 
4x

55. OR
P  x   x 3  19 x  30
 ( x  2)  x 2 2x 15 
constant  constant
=constant
leading term  leading term
=leading term
If you where to expand out now, how many x would you have? 15x
How many x do you need?
19x
How do you get from what you have to what you need?
4 x  2  ?
 P ( x)  ( x  2)  x 2  2 x  15 
 ( x  2)( x  5)( x  3)
4x

56. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36

57. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant

58. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

59. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!

60. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
Fractional factors must be of the form

61. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient

62. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4

63. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4

64. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2

65. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2

66. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too

67. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
3
2

68. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
0
3
2

69. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
0
 ( x  4) is a factor
3
2

70. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
0
 ( x  4) is a factor
3
2
P( x)  4 x 3  16 x 2  9 x  36
  x  4 


71. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
0
 ( x  4) is a factor
3
2
P( x)  4 x 3  16 x 2  9 x  36
  x  4  4x 2


72. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
0
 ( x  4) is a factor
3
2
P( x)  4 x 3  16 x 2  9 x  36
  x  4  4x 2
9 

73. (ii) Factorise P  x   4 x 3  16 x 2  9 x  36
Constant factors must be a factor of the constant
Possibilities = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36
of course they could be negative!!!
factors of the constant
Fractional factors must be of the form
factors of the leading coefficient
1 2 3 4 6 9 12 18 36
Possibilities = , , , , , , , ,
4 4 4 4 4 4 4 4 4
1 2 3 4 6 9 12 18 36
= , , , , , , , ,
2 2 2 2 2 2 2 2 2 they could be negative too
P  4   4  4   16  4   9  4   36
0
 ( x  4) is a factor
3
2
P( x)  4 x 3  16 x 2  9 x  36
  x  4  4x 2
9 
  x  4  2 x  3 2 x  3

74. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?

75. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11

76. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11

77. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11
(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?

78. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11
(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?
P3  1

79. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11
(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?
P3  1
4a  b  1

80. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11
(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?
P3  1
4a  b  1
4a  12
a3

81. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11
(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?
P3  1
4a  b  1
4a  12
a3
P x    x  1 x  3Q x   3 x  8

82. 2004 Extension 1 HSC Q3b)
Let P(x) = (x + 1)(x – 3)Q(x) + a(x + 1) + b, where Q(x) is a polynomial
and a and b are real numbers.
When P(x) is divided by (x + 1) the remainder is – 11. When P(x) is
divided by (x – 3) the remainder is 1.
(i) What is the value of b?
P 1  11
b  11
(ii) What is the remainder when P(x) is divided by (x + 1)(x – 3)?
P3  1
4a  b  1
4a  12
a3
P x    x  1 x  3Q x   3 x  8
 R x   3x  8

83. 2002 Extension 1 HSC Q2c)
Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial.
Find the value of a.

84. 2002 Extension 1 HSC Q2c)
Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial.
Find the value of a.
P  2   3

85. 2002 Extension 1 HSC Q2c)
Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial.
Find the value of a.
P  2   3
 23  2 22  a  3

86. 2002 Extension 1 HSC Q2c)
Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial.
Find the value of a.
P  2   3
 23  2 22  a  3
 16  a  3

87. 2002 Extension 1 HSC Q2c)
Suppose x 3  2 x 2  a   x  2 Q x   3 where Q(x) is a polynomial.
Find the value of a.
P  2   3
 23  2 22  a  3
 16  a  3
a  19

88. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?

89. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.

90. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5

91. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)

92. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)

93. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5

94. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)

95. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
R4  5

96. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
R4  5
4a  b  5

97. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
P 1  5
R4  5
4a  b  5

98. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
P 1  5
R4  5
ab 5
4a  b  5

99. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
P 1  5
R4  5
ab 5
4a  b  5
 5a  10
a  2

100. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
P 1  5
R4  5
ab 5
4a  b  5
 5a  10
a  2 b  3

101. 1994 Extension 1 HSC Q4a)
When the polynomial P(x) is divided by (x + 1)(x – 4), the quotient is
Q(x) and the remainder is R(x).
(i) Why is the most general form of R(x) given by R(x) = ax + b?
The degree of the divisor is 2, therefore the degree of the
remainder is at most 1, i.e. a linear function.
(ii) Given that P(4) = – 5 , show that R(4) = – 5
P(x) = (x + 1)(x – 4)Q(x) + R(x)
P(4) = (4 + 1)(4 – 4)Q(4) + R(4)
R(4) = – 5
(iii) Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x)
P 1  5
R4  5
ab 5
4a  b  5
 5a  10
R x   2 x  3
a  2 b  3

102. 2x 1
1
use P  
2
Exercise 4D; 1bc, 3ac, 4ac, 7acf, 8bdf, 9b, 12ac,
13, 14, 17, 23*