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Fundamental Theorem The Fundamental Theorem of Algebra

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9/2/2013 Fundamental Theorem 2 Quadratic Formula Examples 1. Solve: x 2 – 2x + 1 = 0 The Fundamental Theorem x = 2a2a b 2 – 4 ac ± –b–b = 1 = 2 0 ± 2 Solution set: { 1 } Question: Is there an easier way ? = 2(1) (–2) 2 – 4(1)(1) ± +2+2

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9/2/2013 Fundamental Theorem 3 Quadratic Formula Examples 2. Solve: The Fundamental Theorem x = 2a2a b 2 – 4 ac ± –b–b = 2 36 ± 4 Solution set: { –1, 5 } Question: Is there an easier way ? x 2 – 4x – 5 = 0 = 2(1) (–4) 2 – 4(1)(–5) ± +4+4 x = 5 –1

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9/2/2013 Fundamental Theorem 4 Quadratic Formula Examples 3. Solve: The Fundamental Theorem x = 2a2a b 2 – 4 ac ± –b–b = 2 –16 ± 2 Solution set: Question: Is there an easier way ? = (–2) 2 – 4(1)(5) 2(1) ± +2+2 x x 2 – 2x + 5 = 0 = 1 ± 2 i { 1 – 2 i, 1 + 2 i }

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9/2/2013 Fundamental Theorem 5 Complex Solutions Complex solutions always occur in conjugate pairs Depends on discriminant of the quadratic formula : The Fundamental Theorem WHY ? Note ± radical x = 2a2a b 2 – 4 ac ± –b–b

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9/2/2013 Fundamental Theorem 6 Complex Solutions Quadratic formula : The Fundamental Theorem = b 2 – 4 ac 0 0 > One real solution Two real solutions Two complex conjugate x = 2a2a b 2 – 4 ac ± –b–b 0 < solutions Discriminant

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9/2/2013 Fundamental Theorem 7 The Fundamental Theorem of Algebra A polynomial of degree n ≥ 1 has at least one complex zero Consequence: Every polynomial has a complete factorization The Fundamental Theorem WHY ? For polynomial f(x) there exists a zero k 1 such that f(x) = (x – k 1 )Q 1 (x)

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9/2/2013 Fundamental Theorem 8 For polynomial f(x) there exists a zero k 1 such that f(x) = (x – k 1 )Q 1 (x) The Fundamental Theorem where deg Q 1 (x) = deg f(x) – 1 Apply the Fundamental Theorem to Q 1 (x) Q 1 (x) = (x – k 2 )Q 2 (x) so f(x) = (x – k 1 )(x – k 2 )Q 2 (x) where deg Q 2 (x) = deg f(x) – 2

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9/2/2013 Fundamental Theorem 9 The Fundamental Theorem f(x) = (x – k 1 )(x – k 2 )Q 2 (x) where deg Q 2 (x) = deg f(x) – 2 For deg f(x) = n, apply n times : f(x) is now completely factored ! f(x) = (x – k 1 )(x – k 2 ) (x – k n )C n where C n is a nonzero constant

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9/2/2013 Fundamental Theorem 10 The Number of Zeros Theorem A polynomial of degree n has at most n distinct zeros From the Complete Factorization: f(x) = (x – k 1 )(x – k 2 )...(x – k n )C n If all k i are distinct there are n distinct zeros... otherwise one or more zeros are repeated zeros The Fundamental Theorem

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9/2/2013 Fundamental Theorem 11 The Number of Zeros Theorem Examples 1. f(x) = 5x(x + 3)(x – 1)(x – 4) 2. f(x) = 4x 2 (x – 2) 3 (x – 7) The Fundamental Theorem How many distinct zeros ? Total zeros ?

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9/2/2013 Fundamental Theorem 12 The Conjugate Zeros Theorem If f(x) is a polynomial with only real coefficients then its complex zeros occur in conjugate pairs Examples 1. f(x) = x 3 + x = x(x 2 + 1) = x(x + i )(x – i ) The Fundamental Theorem Zeros ?

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9/2/2013 Fundamental Theorem 13 The Conjugate Zeros Theorem If f(x) is a polynomial with only real coefficients then its complex zeros occur in conjugate pairs Examples 2. f(x) = x 4 + 5x 2 – 36 = u 2 + 5u – 36 = (u – 4)(u + 9) = (x 2 – 4)(x 2 + 9) The Fundamental Theorem Let u = x 2 = (x + 2)(x – 2)(x + 3 i )(x – 3 i )Zeros ?

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9/2/2013 Fundamental Theorem 14 Solving Equations Using Zeros Factor polynomial Set completely factored form equal to 0 Apply zero product property Examples 1. f(x) = x 4 + 5x 2 – 36 The Fundamental Theorem = (x + 2)(x – 2)(x + 3 i )(x – 3 i ) = 0 OR Solution set: { –2, 2, –3 i, 3 i } x + 2 = 0ThusOR x – 2 = 0 OR x + 3 i = 0 x – 3 i = 0

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9/2/2013 Fundamental Theorem 15 Solving Equations Using Zeros Factor polynomial Set completely factored form equal to 0 Apply zero product property Examples 2. f(x) = x 3 + x The Fundamental Theorem = 0 x + i = 0 Thus OR x – i = 0 x = 0 OR Solution set: { 0, – i, i } = x (x + i )(x – i )

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9/2/2013 Fundamental Theorem 16 Think about it !

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9/2/2013 Fundamental Theorem 17 x 2 = –1 i Spare Parts ± ± –1 = ± x i = –i–i – = – a – a

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