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Complex Zeros; Fundamental Theorem of Algebra
Objective:
โ€ข SWBAT identify complex zeros of a polynomials by using
Conjugate Root Theorem
โ€ข SWBAT find all real and complex zeros by using
Fundamental Theorem of Algebra
Complex Zeros; Fundamental Theorem of Algebra
Complex Numbers
The complex number system includes real and imaginary
numbers.
Standard form of a complex number is: a + bi.
a and b are real numbers.
i is the imaginary unit โˆ’1 ๏‚ฎ (๐‘– 2 = โˆ’1).
Fundamental Theorem of Algebra
Every complex polynomial function of degree 1 or larger
(no negative integers as exponents) has at least one
complex zero.
Complex Zeros; Fundamental Theorem of Algebra
Theorem
Every complex polynomial function of degree n ๏‚ณ 1 has
exactly n complex zeros, some of which may repeat.
Conjugate Pairs Theorem
If ๐‘Ÿ = ๐‘Ž + ๐‘๐‘– is a zero of a polynomial function whose
coefficients are real numbers, then the complex
conjugate ๐‘Ÿ = ๐‘Ž โˆ’ ๐‘๐‘– is also a zero of the function.
Examples
1) A polynomial function of degree three has 2 and 3 + i
as it zeros. What is the other zero?
๐‘ฅ =3โˆ’๐‘–
Complex Zeros; Fundamental Theorem of Algebra
Examples
2) A polynomial function of degree 5 has 4, 2 + 3i, and 5i
as it zeros. What are the other zeros?
๐‘ฅ = 2 โˆ’ 3๐‘–
๐‘Ž๐‘›๐‘‘
๐‘ฅ = โˆ’5๐‘–
3) A polynomial function of degree 4 has 2 with a zero
multiplicity of 2 and 2 โ€“ i as it zeros. What are the
zeros?
๐‘Ž๐‘›๐‘‘ ๐‘ฅ = 2 + ๐‘–
๐‘ฅ = 2 ๐‘Ÿ๐‘’๐‘๐‘’๐‘Ž๐‘ก๐‘  ๐‘ก๐‘ค๐‘–๐‘๐‘’
Complex Zeros; Fundamental Theorem of Algebra
Examples
4) A polynomial function of degree 4 has 2 with a zero
multiplicity of 2 and 2 โ€“ i as it zeros. What is the function?
๐‘ฅ=2
๐‘ฅ=2
๐‘ฅ =2โˆ’๐‘–
๐‘ฅ =2+๐‘–
๐‘“ ๐‘ฅ = (๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ (2 โˆ’ ๐‘–))(๐‘ฅ โˆ’ (2 + ๐‘–))
๐‘“ ๐‘ฅ = (๐‘ฅ 2 โˆ’4๐‘ฅ + 4)(๐‘ฅ โˆ’ 2 + ๐‘–)(๐‘ฅ โˆ’ 2 โˆ’ ๐‘–)
(๐‘ฅ 2 โˆ’ 2๐‘ฅ โˆ’ ๐‘–๐‘ฅ โˆ’ 2๐‘ฅ + 4 + 2๐‘– + ๐‘–๐‘ฅ โˆ’ 2๐‘– โˆ’ ๐‘– 2 )
๐‘“ ๐‘ฅ = (๐‘ฅ 2 โˆ’4๐‘ฅ + 4)(๐‘ฅ 2 โˆ’ 4๐‘ฅ + 5)
๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’ 4๐‘ฅ 3 + 5๐‘ฅ 2 โˆ’ 4๐‘ฅ 3 + 16๐‘ฅ 2 โˆ’ 20๐‘ฅ + 4๐‘ฅ 2 โˆ’ 16๐‘ฅ + 20
๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’ 8๐‘ฅ 3 + 25๐‘ฅ 2 โˆ’ 36๐‘ฅ + 20
Complex Zeros; Fundamental Theorem of Algebra
Find the remaining complex zeros of the given polynomial
functions
5) ๐‘“ ๐‘ฅ = ๐‘ฅ 3 + 3๐‘ฅ 2 + 25๐‘ฅ + 75 ๐‘ง๐‘’๐‘Ÿ๐‘œ: โˆ’5๐‘–
๐ด๐‘›๐‘œ๐‘กโ„Ž๐‘’๐‘Ÿ ๐‘ง๐‘’๐‘Ÿ๐‘œ (๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘›๐‘—๐‘ข๐‘”๐‘Ž๐‘ก๐‘’): 5๐‘–
๐‘ฅ = โˆ’5๐‘– ๐‘Ž๐‘›๐‘‘ ๐‘ฅ = 5๐‘–
(๐‘ฅ + 5๐‘–)(๐‘ฅ โˆ’ 5๐‘–)
๐‘ฅ 2 โˆ’ 5๐‘–๐‘ฅ + 5๐‘–๐‘ฅ โˆ’ 25๐‘– 2
๐‘ฅ 2 โˆ’ 25(โˆ’1)
๐‘ฅ 2 + 25
Complex Zeros; Fundamental Theorem of Algebra
๐‘“ ๐‘ฅ = ๐‘ฅ 3 + 3๐‘ฅ 2 + 25๐‘ฅ + 75
๐‘ง๐‘’๐‘Ÿ๐‘œ: โˆ’5๐‘–
Long Division
๐‘ฅ +3
x ๏€ซ 25 x ๏€ซ 3 x ๏€ซ 25 x ๏€ซ 75
2
3
2
๐‘ฅ3
25๐‘ฅ
3๐‘ฅ 22
3๐‘ฅ
+75
+75
0
๐‘ฅ + 3 ๐‘ฅ 2 + 25
(๐‘ฅ + 3)(๐‘ฅ + 5๐‘–)(๐‘ฅ โˆ’ 5๐‘–)
๐‘ง๐‘’๐‘Ÿ๐‘œ๐‘ : โˆ’3, โˆ’5๐‘– ๐‘Ž๐‘›๐‘‘ 5๐‘–
Complex Zeros; Fundamental Theorem of Algebra
Find the complex zeros of the given polynomial functions
6) ๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’ 4๐‘ฅ 3 + 9๐‘ฅ 2 โˆ’ 20๐‘ฅ + 20
๐‘: ±1, ±2, ±4, ±5, ±10, ±20
๐‘ž: ±1
๐‘
1 2 4 5 10 20
: ± ,± ,± ,± ,± ,±
๐‘ž
1 1 1 1
1
1
Possible solutions: ๐‘ฅ = ±1, ±2, ±4, ±5, ±10, ±20
Try: ๐‘ฅ = โˆ’1
Try: ๐‘ฅ = 1
1 1 ๏€ญ 4 9 ๏€ญ 20 20
1
6 โˆ’14
1 โˆ’3
โˆ’3 6 โˆ’14 6
๏€ญ 1 1 ๏€ญ 4 9 ๏€ญ 20 20
1
โˆ’1 5 โˆ’14
โˆ’5 14 โˆ’34
34
54
Complex Zeros; Fundamental Theorem of Algebra
๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’ 4๐‘ฅ 3 + 9๐‘ฅ 2 โˆ’ 20๐‘ฅ + 20
Try: ๐‘ฅ = 2
2 1 ๏€ญ 4 9 ๏€ญ 20 20
10 โˆ’20
2 โˆ’4
1 โˆ’2 5 โˆ’10 0
๐‘“ ๐‘ฅ = (๐‘ฅ โˆ’ 2)(๐‘ฅ 3 โˆ’ 2๐‘ฅ 2 + 5๐‘ฅ โˆ’ 10)
๐‘“ ๐‘ฅ = (๐‘ฅ โˆ’ 2)(๐‘ฅ 2 ๐‘ฅ โˆ’ 2 + 5(๐‘ฅ โˆ’ 2))
๐‘“ ๐‘ฅ = (๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ 2)(๐‘ฅ 2 + 5)
Complex Zeros; Fundamental Theorem of Algebra
๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’ 4๐‘ฅ 3 + 9๐‘ฅ 2 โˆ’ 20๐‘ฅ + 20
๐‘“ ๐‘ฅ = ๐‘ฅ โˆ’ 2 ๐‘ฅ โˆ’ 2 ๐‘ฅ2 + 5 = 0
๐‘ฅ=2
๐‘ฅ2 + 5 = 0
๐‘ฅ 2 = โˆ’5
๐‘ง๐‘’๐‘Ÿ๐‘œ ๐‘š๐‘ข๐‘™๐‘ก๐‘–๐‘๐‘™๐‘–๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ 2
๐‘ฅ = ± โˆ’5
๐‘ฅโˆ’2=0
๐‘ฅโˆ’2=0
๐‘ฅ =± 5๐‘–
Complex zeros: 2 with multiplicity of 2, 5 ๐‘–, ๐‘Ž๐‘›๐‘‘ โˆ’ 5 ๐‘–
๐‘“ ๐‘ฅ ๐‘–๐‘› ๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘’๐‘‘ ๐‘“๐‘œ๐‘Ÿ๐‘š
๐‘“ ๐‘ฅ = (๐‘ฅ โˆ’ 2)2 ๐‘ฅ โˆ’ 5 ๐‘– ๐‘ฅ + 5 ๐‘–
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