Download presentation

Presentation is loading. Please wait.

1
**Complex Zeros of Polynomial**

By: Mao & Na

2
**Introduction: What you'll learn about : Two Major Theorems**

Complex Conjugate Zeros Factoring with Real Number Coefficients.

3
Video ftI8k

4
Two Major Theorems Fundamental Theorem of Algebra: A polynomial function of degree n has n complex zeros( real or nonreal). Some of these zeros may be repeated. Liner Factorization Theorem: If f(x) is a polynomial function of degree n >0, then f(x) has precisely n linear factors and f(x) = a(x –z1)(x –z2) (x -zn) Where a is the leading coefficient of f(x) and z1, z2,….., zn are the complex zeros of f(x). The zi are not necessarily distinct numbers; some may be repeated.

5
**Fundamental Polynomial Connections in the Complex Case:**

The following statements about a polynomial function f are equivalent if k is a complex number. 1. x = k is a solution (or root) of the equation f(x) = 0 2. is a zero of the function f. 3.x - k is a factor of f(x).

6
**Complex Conjugate Zeros**

Suppose that f(x) is a polynomial function with real coefficients. If a and b are real numbers with b not equal 0 and a + bi is a zero of f(x), then its complex conjugate a - bi is also a zero of f(x).

7
**Factoring with Real Number Coefficients**

Let f(s) be a polynomial function with real coefficients. The Linear Factorization Theorem tells us that f(x) can be factored into the form f(x) = a(x –z1)(x –z2) (x -zn), Where zi are complex numbers. Recall, however, that nonreal complex zeros occur in conjugate pairs. The product of x - (a + bi) and x - (a - bi) is [x-(a+bi)][x-(a-bi)] = x^(2) - (a-bi)x - (a+bi)x + (a+bi)(a-bi) = x^(2 ) - 2ax + ((a^2) + (b^2)).

8
**Polynomial Function of Odd Degree**

Factoring with Real Number Coefficients Every polynomial function with real coefficients can be written as a product of linear factors and irreducible quadratic factors, each with real coefficients. Polynomial Function of Odd Degree Every polynomial function of odd degree with real coefficients has at least one real zero.

9
**Complex Zeros of Polynomial**

Find complex zeros of polynomial function : F(x) = 3x^(4) + 5x^(3) + 25x^(2) + 45x – 18

10
/// / /

11
Assessment of Content Write the polynomial in standard from, and identify the zeros of the function and the x-intercept of it’s graph. 1.f(x)=(x-3i)(x+3i) A. x2+9; zeros:+-3i: x-intercepts: none B. x2+6; zeros:+-3i: x-intercepts: 2 2.f(x)=(x-1)(x-1)(x+2i)(x-2i) A. X4-3x+25x-4x+4 B. X4-2x+5x-8x+4 Write the polynomial function of minimum degree in standard form with real coefficient whose zeros included those listed. 3. I and – I A. x3+4 B. x2+1 4. 2,3 and I A. X4-5x3+7x-5x+6 B. X2-2x2+7x-3x+4 State how many complex and real zeros the function has. 5. f(x)=x2-2x+7 A. 2 complex zeros: none real B. 4 complex zeros: 7 real 6. F(x)=x4-5x3+x2-3x+6

12
**Assessment of Content continue**

Find all of the zeros and write a linear factorization of the function. 7. f(x)= x3+4x-5 A. Zeros: x=1, x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √19i) (2x+1- √ 19i) B. Zeros: x=2, x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √19i) (2x+1- √ 19i) 8. f(x)= x4+x3+5x2-x-6 A. Zeros: x= ± 1, x = -1/2 ± √ 23/2i ; f(x) = ¼ (x-1)(x+1)(2x+1+ √ 19i) (2x+1- √ 23i B. . Zeros: x= ± 1, x = -1/2 ± √ 23/2i ; f(x) = ¼ (x-1)(x+1)(2x+1+ √ 19i) (2x+1- √ 23i Using the given zeros, find all of the zeros and write a linear factorization of f(x). 9. 1+i is a zeros of f(x)=x4-2x3-x2+6x-6 A. Zeros: x=2 x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(3x+1+ √ 19i) (3x+1- √ 19i) B. Zeros: x=1, x = -1/2 ± √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √ 19i) (2x+1- √ 19i) i is a zeros of f(x)=x4-6x3+11x2+12x-26 A. Zeros: x +- √ 2. x =3 ± 2i: f(x) = (x- √ 2) ( x+ √ 2)(x-3+2i)(x-3-2i). B. Zeros: x +- √ 4. x =3 ± 3i: f(x) = (x- √ 2) ( x+ √ 2)(x-3+2i)(x-3-2i).

13
**Answer Key x2+9; zeros:+-3i: x-intercepts: none X4-2x+5x-8x+4 x2+1**

2 complex zeros: none real 4 complex zeros: 2 real Zeros: x=1, x = -1/2+- √ 19/2 i ; f(x) = ¼ (x-1)(2x+1+ √19i) (2x+1- √ 19i) Zeros: x= ± 1, x = -1/2+- square root 23/2i ; f(x) = ¼ (x-1)(x+1)(2x+1+square root 19i) (2x+1- square root 23i) Zeros: x +- √ 3. x =1 +- i: f(x) = (x- √ 3) ( x+ square 3)(x-1+i)(x-1-i). Zeros: x +- √ 2. x =3 +- 2i: f(x) = (x- √ 2) ( x+ √ 2)(x-3+2i)(x-3-2i).

14
Works Cited Demana, Franklin. Waits, Bert K.. Foley, Gregory D.. Kennedy, Daniel. Pre- Calculus. Eight Edition. Graphical, Numerical, Algebraic. "Finding Complex Zeros of a Polynomial Function." YouTube. YouTube, 10 Oct Web. 21 Jan "X Finds Out His Value." YouTube. YouTube, 31 Dec Web. 21 Jan

Similar presentations

OK

1 Copyright © 2015, 2011, and 2007 Pearson Education, Inc. Start Up Day 13.

1 Copyright © 2015, 2011, and 2007 Pearson Education, Inc. Start Up Day 13.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on pre ignition problem Ppt on ms excel formulas Ppt on basics of ms word 2007 Ppt on next generation 2-stroke engine model Unlock ppt online free Ppt on ac to dc converter Download ppt on sources of energy for class 10th Ppt on object oriented programming with c++ Ppt on heritage of indian culture Ppt on light dependent resistor