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Complex Zeros; Fundamental Theorem of Algebra

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Presentation on theme: "Complex Zeros; Fundamental Theorem of Algebra"โ€” Presentation transcript:

1 Complex Zeros; Fundamental Theorem of Algebra
Objective: SWBAT identify complex zeros of a polynomials by using Conjugate Root Theorem SWBAT find all real and complex zeros by using Fundamental Theorem of Algebra

2 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.

3 Complex Zeros; Fundamental Theorem of Algebra
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‘Žโˆ’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‘–

4 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=2โˆ’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‘ฅ=2 ๐‘Ÿ๐‘’๐‘๐‘’๐‘Ž๐‘ก๐‘  ๐‘ก๐‘ค๐‘–๐‘๐‘’ ๐‘Ž๐‘›๐‘‘ ๐‘ฅ=2+๐‘–

5 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‘ฅโˆ’2)(๐‘ฅโˆ’2)(๐‘ฅโˆ’(2โˆ’๐‘–))(๐‘ฅโˆ’(2+๐‘–)) ๐‘“ ๐‘ฅ = (๐‘ฅ 2 โˆ’4๐‘ฅ+4)(๐‘ฅโˆ’2+๐‘–)(๐‘ฅโˆ’2โˆ’๐‘–) ( ๐‘ฅ 2 โˆ’2๐‘ฅโˆ’๐‘–๐‘ฅโˆ’2๐‘ฅ+4+2๐‘–+๐‘–๐‘ฅโˆ’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’4 ๐‘ฅ 3 +5 ๐‘ฅ 2 โˆ’4 ๐‘ฅ ๐‘ฅ 2 โˆ’20๐‘ฅ+4 ๐‘ฅ 2 โˆ’16๐‘ฅ+20 ๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’8 ๐‘ฅ ๐‘ฅ 2 โˆ’36๐‘ฅ+20

6 Complex Zeros; Fundamental Theorem of Algebra
Find the remaining complex zeros of the given polynomial functions 5) ๐‘“ ๐‘ฅ = ๐‘ฅ 3 +3 ๐‘ฅ 2 +25๐‘ฅ 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‘ฅโˆ’5๐‘–) ๐‘ฅ 2 โˆ’5๐‘–๐‘ฅ+5๐‘–๐‘ฅโˆ’25 ๐‘– 2 ๐‘ฅ 2 โˆ’25(โˆ’1) ๐‘ฅ 2 +25

7 Complex Zeros; Fundamental Theorem of Algebra
๐‘“ ๐‘ฅ = ๐‘ฅ 3 +3 ๐‘ฅ 2 +25๐‘ฅ ๐‘ง๐‘’๐‘Ÿ๐‘œ:โˆ’5๐‘– Long Division ๐‘ฅ +3 ๐‘ฅ 3 25๐‘ฅ 3๐‘ฅ 2 +75 3๐‘ฅ 2 +75 ๐‘ฅ+3 ๐‘ฅ 2 +25 (๐‘ฅ+3)(๐‘ฅ+5๐‘–)(๐‘ฅโˆ’5๐‘–) ๐‘ง๐‘’๐‘Ÿ๐‘œ๐‘ :โˆ’3,โˆ’5๐‘– ๐‘Ž๐‘›๐‘‘ 5๐‘–

8 Complex Zeros; Fundamental Theorem of Algebra
Find the complex zeros of the given polynomial functions 6) ๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’4 ๐‘ฅ 3 +9 ๐‘ฅ 2 โˆ’20๐‘ฅ+20 ๐‘: ยฑ1, ยฑ2, ยฑ4, ยฑ5, ยฑ10, ยฑ ๐‘ž: ยฑ1 ๐‘ ๐‘ž : ยฑ 1 1 , ยฑ 2 1 , ยฑ 4 1 , ยฑ 5 1 , ยฑ 10 1 , ยฑ 20 1 Possible solutions: ๐‘ฅ=ยฑ1, ยฑ2, ยฑ4, ยฑ5,ยฑ10,ยฑ20 Try: ๐‘ฅ=1 Try: ๐‘ฅ=โˆ’1 1 โˆ’3 6 โˆ’14 โˆ’1 5 โˆ’14 34 1 โˆ’3 6 โˆ’14 6 1 โˆ’5 14 โˆ’34 54

9 Complex Zeros; Fundamental Theorem of Algebra
๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’4 ๐‘ฅ 3 +9 ๐‘ฅ 2 โˆ’20๐‘ฅ+20 Try: ๐‘ฅ=2 2 โˆ’4 10 โˆ’20 1 โˆ’2 5 โˆ’10 ๐‘“ ๐‘ฅ =(๐‘ฅโˆ’2)( ๐‘ฅ 3 โˆ’2 ๐‘ฅ 2 +5๐‘ฅโˆ’10) ๐‘“ ๐‘ฅ =(๐‘ฅโˆ’2)( ๐‘ฅ 2 ๐‘ฅโˆ’2 +5(๐‘ฅโˆ’2)) ๐‘“ ๐‘ฅ =(๐‘ฅโˆ’2)(๐‘ฅโˆ’2)( ๐‘ฅ 2 +5)

10 Complex Zeros; Fundamental Theorem of Algebra
๐‘“ ๐‘ฅ = ๐‘ฅ 4 โˆ’4 ๐‘ฅ 3 +9 ๐‘ฅ 2 โˆ’20๐‘ฅ+20 ๐‘“ ๐‘ฅ = ๐‘ฅโˆ’2 ๐‘ฅโˆ’2 ๐‘ฅ 2 +5 =0 ๐‘ฅโˆ’2=0 ๐‘ฅโˆ’2=0 ๐‘ฅ 2 +5=0 ๐‘ฅ 2 =โˆ’5 ๐‘ฅ=2 ๐‘ฅ=ยฑ 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=ยฑ 5 ๐‘– 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‘ฅโˆ’2) 2 ๐‘ฅโˆ’ 5 ๐‘– ๐‘ฅ+ 5 ๐‘–


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