Download presentation

Presentation is loading. Please wait.

Published byEllis Tirey Modified over 2 years ago

1
Polynomial Zeros Real Zeros of Polynomial Functions

2
7/23/2013 Polynomial Zeros 2 Even/Odd Multiplicity Examples Polynomial Functions x y(x) x x x x y = (x – 3) 2 y = x + 3 y = (x – 3) 3 y = (x – 3) 4 y = (x – 3) 5 y = (x + 3) 3 (x – 3) x y(x) y = (x + 2) 3 (x – 3) 2

3
7/23/2013 Polynomial Zeros 3 Rational Zero Test Let f(x) = a n x n + … + a 2 x 2 + a 1 x + a 0 for integer coefficients with a n 0 Then all rational zeros of f(x) are of form p/q p is a factor of a 0 and q is a factor of a n p and q have no common factors Polynomial Functions

4
7/23/2013 Polynomial Zeros 4 Rational Zero Test f(x) = a n x n + … + a 2 x 2 + a 1 x + a 0 A ll rational zeros of f(x) of form p/q, with p is a factor of a 0 and q is a factor of a n NOTE: This works only for integer coefficients NOT all zeros are rational numbers NO irrational zeros of f(x) are included Polynomial Functions

5
7/23/2013 Polynomial Zeros 5 Zeros and Factors FACT: If two polynomials are equal then they have the same factors If f(x) = (x – k 1 )Q 1 (x) and if Q 1 (x) = (x – k 2 )Q 2 (x) then we have f(x) = (x – k 1 )(x – k 2 )Q 2 (x) Polynomial Functions

6
7/23/2013 Polynomial Zeros 6 Rational Zero Test Example Factor completely: f(x) = 3x 4 – 12x 3 – 24x 2 + 36x + 45 = 3(x 4 – 4x 3 – 8x 2 + 12x + 15) = 3g(x) Here a n = a 4 = 1 and a 0 = 15 Factors p of 15 are: ±1, ±3, ±5, ±15 ; Factors q of 1 are: ±1 Possible p/q values are: ±1, ±3, ±5, ±15 Polynomial Functions

7
7/23/2013 Polynomial Zeros 7 Rational Zero Test Example g(x) = x 4 – 4x 3 – 8x 2 + 12x + 15 Possible p/q values: ±1, ±3, ±5, ±15 Check zeros of g(x) Polynomial Functions 1 1 –4 –8 12 15 1 1 –3 –11 16 1 1 k = 1 (x – 1) is NOT a factor of g(x) 1 –4 –8 12 15 1 3 –1 –3 –11 –33 –21 k = 3 –63 –48 (x – 3) is NOT a factor of g(x) 3

8
7/23/2013 Polynomial Zeros 8 Rational Zero Test Example Possible p/q values: ±1, ±3, ±5, ±15 Polynomial Functions 5 1 –4 –8 12 15 1 5 1 5 –3 –15 0 –3 –15 k = 5 (x – 5) IS a factor of g(x) 1 1 –3 –3 1 –1 0 0 –3 3 0 k = –1 (x + 1) IS a factor of Q 1 (x) –1 Q 1 (x) Q 2 (x)

9
7/23/2013 Polynomial Zeros 9 Rational Zero Test Example Polynomial Functions Thus far f(x) = 3g(x) = 3(x – 5)Q 1 (x) = 3(x – 5)(x + 1)Q 2 (x) = 3(x – 5)(x + 1)(x 2 – 3) Q 2 (x) = x 2 – 3 with possible rational zeros of ±1 and ±3 Synthetic division shows none of these are zeros of Q 2 (x)

10
7/23/2013 Polynomial Zeros 10 Rational Zero Test Example Polynomial Functions Q 2 (x) = x 2 – 3 = 3 x 2 – ( ) 2 = 3 ( x – ) 3 ( x + ) f(x)= 3(x – 5)(x + 1)Q 2 (x) Thus 3 ( x – ) 3 ( x + ) = 3(x – 5)(x + 1) Two rational zeros and two irrational zeros 3 x = 5, –1, ±

11
7/23/2013 Polynomial Zeros 11 Equations New functions f(x) lead to new types of equations to solve Set f(x) = 0 and find the zeros Examples Find all real solutions of: 1. x 4 – 1 = 0 Polynomial Functions

12
7/23/2013 Polynomial Zeros 12 Equations Examples Find all real solutions of: 2. x 3 = x 3. x 4 – 5x 2 + 4 = 0 4. x 6 – 19x 3 = 216 Polynomial Functions

13
7/23/2013 Polynomial Zeros 13 Problem Solving Find the width W of the rectangle from its length and area A. Also determine W when L = 5 inches. A = WL = W(x 2 + 1) = 3x 3 + 3x – 5x 2 – 5 = 3x(x 2 + 1) – 5(x 2 + 1) = (3x – 5)(x 2 + 1) Polynomial Functions W A = 3x 3 – 5x 2 + 3x – 5 L = x 2 + 1 A

14
7/23/2013 Polynomial Zeros 14 Problem Solving A = (3x – 5)(x 2 + 1) Polynomial Functions W A = 3x 3 – 5x 2 + 3x – 5 L = x 2 + 1 W = x 2 + 1 A = (3x – 5)(x 2 + 1) = 3x – 5 At x = 5 inches, W = 3(5) – 5 = 10 inches Note:To find W we could have used conventional long division

15
7/23/2013 Polynomial Zeros 15 Think about it !

16
7/23/2013 Polynomial Zeros 16 Polynomial Functions x 2 + 1 W = L A To find W we could have used conventional long division 3x 3 – 5x 2 + 3x – 5 x 2 + 1 = L = x 2 + 1 Note: 3x 3 – 5x 2 + 3x – 5 x 2 + 1

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google