The document discusses factorizing complex expressions. It states that if a polynomial's coefficients are real, its roots will appear in complex conjugate pairs. Any polynomial of degree n can be factorized into a mixture of quadratic and linear factors over real numbers or into n linear factors over complex numbers. Odd degree polynomials must have at least one real root. Examples of factorizing polynomials over both real and complex numbers are provided.
2. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
3. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
4. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
5. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
6. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
7. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
e.g . i x 2 2 x 2
8. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
e.g . i x 2 2 x 2 x 12 1
9. Factorising Complex
Expressions
If a polynomial’s coefficients are all real then the roots will appear
in complex conjugate pairs.
Every polynomial of degree n can be;
• factorised as a mixture of quadratic and linear factors over the
real field
• factorised to n linear factors over the complex field
NOTE: odd ordered polynomials must have a real root
e.g . i x 2 2 x 2 x 12 1
x 1 i x 1 i
13. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
14. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0
15. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
16. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
17. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
iii Factorise 2 x 3 3x 2 8 x 5
18. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
iii Factorise 2 x 3 3x 2 8 x 5
as it is a cubic it must have a real factor
19. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
iii Factorise 2 x 3 3x 2 8 x 5
as it is a cubic it must have a real factor
3 2
1 1 1 1
P 2 3 8 5
2 2 2 2
1 3
45
4 4
0
20. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
iii Factorise 2 x 3 3x 2 8 x 5
as it is a cubic it must have a real factor
3 2
1 1 1 1
P 2 3 8 5
2 2 2 2
1 3
45
4 4
0
2 x 3 3 x 2 8 x 5 2 x 1x 2 2 x 5
21. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
iii Factorise 2 x 3 3x 2 8 x 5
as it is a cubic it must have a real factor
3 2
1 1 1 1
P 2 3 8 5
2 2 2 2
1 3
45
4 4
0
2 x 3 3 x 2 8 x 5 2 x 1x 2 2 x 5
2 x 1 x 1 4
2
22. ii z 4 z 2 12 0
z 3z 4 0
2 2
z 3 z 3z 4 0
2
factorised over Real numbers
z 3 z 3 z 2i z 2i 0 factorised over Complex numbers
z 3 or z 2i
iii Factorise 2 x 3 3x 2 8 x 5
as it is a cubic it must have a real factor
3 2
1 1 1 1
P 2 3 8 5
2 2 2 2
1 3
45
4 4
0
2 x 3 3 x 2 8 x 5 2 x 1x 2 2 x 5
2 x 1 x 1 4
2
2 x 1 x 1 2i x 1 2i
24. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
25. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
And z 6 1 0 has solutions;
2 k
z cis k 0, 1, 2,3
6
26. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
And z 6 1 0 has solutions;
2 k
z cis k 0, 1, 2,3
6
z5 z4 z3 z2 1 0
27. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
And z 6 1 0 has solutions;
2 k
z cis k 0, 1, 2,3
6
z5 z4 z3 z2 1 0
z 1z 5 z 4 z 3 z 2 1
0
z 1
28. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
And z 6 1 0 has solutions;
2 k
z cis k 0, 1, 2,3
6
z5 z4 z3 z2 1 0
z 1z 5 z 4 z 3 z 2 1
0
z 1
z 6 1 0, z 1
29. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
And z 6 1 0 has solutions;
2 k
z cis k 0, 1, 2,3
6
z5 z4 z3 z2 1 0
z 1z 5 z 4 z 3 z 2 1
0
z 1
z 6 1 0, z 1
2 2
z cis , cis , cis , cis , cis
3 3 3 3
30. iv z 5 z 4 z 3 z 2 z 1 0
Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1
And z 6 1 0 has solutions;
2 k
z cis k 0, 1, 2,3
6
z5 z4 z3 z2 1 0
z 1z 5 z 4 z 3 z 2 1
0
z 1
z 6 1 0, z 1
2 2
z cis , cis , cis , cis , cis
3 3 3 3
1 3 1 3 1 3 1 3
z i, i, i, i, 1
2 2 2 2 2 2 2 2
31. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
32. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
33. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
34. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
35. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
z k
36. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
z k
b) Prove that 2 3 4 5 6 7 8 1
37. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
z k
b) Prove that 2 3 4 5 6 7 8 1
z9 1 0
38. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
z k
b) Prove that 2 3 4 5 6 7 8 1
z9 1 0
1 2 3 4 5 6 7 8 0
39. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
z k
b) Prove that 2 3 4 5 6 7 8 1
z9 1 0
1 2 3 4 5 6 7 8 0 sum of roots
40. v 1996 HSC
2 2
Let cos i sin
9 9
a) Show that k is a solution of z 9 1 0, where k is an integer
z9 1
2k
z cis
9 k 0,1,2,3,4,5,6,7,8
2
k
z cis
9
z k
b) Prove that 2 3 4 5 6 7 8 1
z9 1 0
1 2 3 4 5 6 7 8 0 sum of roots
2 3 4 5 6 7 8 1
41. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
42. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B
sin A sin B 2sin cos
2 2
43. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
44. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
2 2
45. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
46. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B
cos A cos B 2sin sin
2 2
47. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B (minus 2 sine half sum
cos A cos B 2sin sin
2 2 sine half diff)
48. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B (minus 2 sine half sum
cos A cos B 2sin sin
2 2 sine half diff)
2 3 4 5 6 7 8 1
49. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B (minus 2 sine half sum
cos A cos B 2sin sin
2 2 sine half diff)
2 3 4 5 6 7 8 1
roots appear in conjugate pairs
50. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B (minus 2 sine half sum
cos A cos B 2sin sin
2 2 sine half diff)
2 3 4 5 6 7 8 1
roots appear in conjugate pairs
2 4 6 8
2cos 2cos 2cos 2cos 1
9 9 9 9
51. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B (minus 2 sine half sum
cos A cos B 2sin sin
2 2 sine half diff)
2 3 4 5 6 7 8 1
roots appear in conjugate pairs
2 4 6 8
2cos 2cos 2cos 2cos 1
9 9 9 9
2 4 6 8 1
cos cos cos cos
9 9 9 9 2
52. 2 4 1
c) Hence show that cos cos cos
9 9 9 8
A B A B (2 sine half sum
sin A sin B 2sin cos
2 2 cos half diff)
A B cos A B
cos A cos B 2 cos
(2 cos half sum
2 2 cos half diff)
A B A B (minus 2 sine half sum
cos A cos B 2sin sin
2 2 sine half diff)
2 3 4 5 6 7 8 1
roots appear in conjugate pairs
2 4 6 8
2cos 2cos 2cos 2cos 1
9 9 9 9
2 4 6 8 1
cos cos cos cos
9 9 9 9 2
3 7 1
2cos cos 2cos cos
9 9 9 9 2
54. 3 7 1
2cos cos 2cos cos
9 9 9 9 2
3 7 1
cos cos cos
9 9 9 4
55. 3 7 1
2cos cos 2cos cos
9 9 9 9 2
3 7 1
cos cos cos
9 9 9 4
5 2 1
cos 2cos cos
9 9 9 4
56. 3 7 1
2cos cos 2cos cos
9 9 9 9 2
3 7 1
cos cos cos
9 9 9 4
5 2 1
cos 2cos cos
9 9 9 4
2 5 1
cos cos cos
9 9 9 8
57. 3 7 1
2cos cos 2cos cos
9 9 9 9 2
3 7 1
cos cos cos
9 9 9 4
5 2 1
cos 2cos cos
9 9 9 4
2 5 1
cos cos cos
9 9 9 8
5 4
But cos cos
9 9
58. 3 7 1
2cos cos 2cos cos
9 9 9 9 2
3 7 1
cos cos cos
9 9 9 4
5 2 1
cos 2cos cos
9 9 9 4
2 5 1
cos cos cos
9 9 9 8
5 4
But cos cos
9 9
2 4 1
cos cos cos
9 9 9 8
59. 3 7 1
2cos cos 2cos cos
9 9 9 9 2
3 7 1
cos cos cos
9 9 9 4
5 2 1
cos 2cos cos
9 9 9 4
2 5 1
cos cos cos
9 9 9 8
5 4
But cos cos
9 9
2 4 1
cos cos cos
9 9 9 8
2 4 1
cos cos cos
9 9 9 8
60. OR
z9 1
z 1 z z 8 z 2 z 7 z 3 z 6 z 4 z 5
61. OR
z9 1
z 1 z z 8 z 2 z 7 z 3 z 6 z 4 z 5
2 4
z 1 z 2 2cos z 1 z 2 2cos z 1
9 9
2 6 2 8
z 2cos z 1 z 2cos z 1
9 9
62. OR
z9 1
z 1 z z 8 z 2 z 7 z 3 z 6 z 4 z 5
2 4
z 1 z 2 2cos z 1 z 2 2cos z 1
9 9
2 6 2 8
z 2cos z 1 z 2cos z 1
9 9
2 2 2 4
z 1 z 2cos z 1 z 2cos z 1
9 9
8
z 2 z 1 z 2 2cos
9
z 1
63. OR
z9 1
z 1 z z 8 z 2 z 7 z 3 z 6 z 4 z 5
2 4
z 1 z 2 2cos z 1 z 2 2cos z 1
9 9
2 6 2 8
z 2cos z 1 z 2cos z 1
9 9
2 2 2 4
z 1 z 2cos z 1 z 2cos z 1
9 9
8
z 2 z 1 z 2 2cos
9
z 1
Let z i
64. OR
z9 1
z 1 z z 8 z 2 z 7 z 3 z 6 z 4 z 5
2 4
z 1 z 2 2cos z 1 z 2 2cos z 1
9 9
2 6 2 8
z 2cos z 1 z 2cos z 1
9 9
2 2 2 4
z 1 z 2cos z 1 z 2cos z 1
9 9
8
z 2 z 1 z 2 2cos
9
z 1
Let z i
2 4 8
i 9 1 i 1 2cos i 2cos i i 2cos i
9 9 9
65. 2 4 8
i 1 i 1 2cos
9
i 2cos i i 2cos i
9 9 9
66. 2 4 8
i 1 i 1 2cos
9
i 2cos i i 2cos i
9 9 9
2 4 8
i 1 i i 1 2cos
4
2cos 2cos
9 9 9
67. 2 4 8
i 1 i 1 2cos
9
i 2cos i i 2cos i
9 9 9
2 4 8
i 1 i i 1 2cos
4
2cos 2cos
9 9 9
2 4 8
1 8cos cos cos
9 9 9
68. 2 4 8
i 1 i 1 2cos
9
i 2cos i i 2cos i
9 9 9
2 4 8
i 1 i i 1 2cos
4
2cos 2cos
9 9 9
2 4 8
1 8cos cos cos
9 9 9
2 4
1 8cos cos cos
9 9 9
69. 2 4 8
i 1 i 1 2cos
9
i 2cos i i 2cos i
9 9 9
2 4 8
i 1 i i 1 2cos
4
2cos 2cos
9 9 9
2 4 8
1 8cos cos cos
9 9 9
2 4
1 8cos cos cos
9 9 9
2 4 1
cos cos cos
9 9 9 8
70. 2 4 8
i 1 i 1 2cos
9
i 2cos i i 2cos i
9 9 9
2 4 8
i 1 i i 1 2cos
4
2cos 2cos
9 9 9
2 4 8
1 8cos cos cos
9 9 9
2 4
1 8cos cos cos
9 9 9
2 4 1
cos cos cos
9 9 9 8
Exercise 4J; 1 to 4, 7ac