4.4 Real Zeros of Polynomial Functions - PowerPoint Presentation

Slide1

4.4 Real Zeros of Polynomial Functions

Understand the factor theorem

Factor higher degree polynomials completely

Analyze polynomials having multiple zeros

Understand the rational zeros test

and Descartes

’ rule of

signs

Solve higher degree polynomial equations

Understand

the intermediate value

theorem

Slide2

Factor Theorem

A polynomial

f

(

x

) has a factor

x

−

k

if and only if

f

(

k

) = 0

.

Slide3

Example: Applying the factor theorem (1 of 2)

Use the graph and the factor theorem to list the factors of

f

(

x

). Assume that all zeros are integers.

Solution

The zeros or

x

-intercepts of

f

are

−

2

, 1 and 3. Since

f

(

−

2

) = 0, the factor theorem states that (

x

+ 2) is a factor, and

f

(1) = 0 implies that (

x

−

1) is a factor and

f

(3) = 0 implies (

x

−

3) is a factor. Thus the factors are (

x

+ 2), (

x

−

1), and

(

x

−

3

).

Slide4

Example: Applying the factor theorem (2 of 2)

Slide5

Zeros with Multiplicity

If f(x) = (x + 2)², then the factor (x + 2) occurs twice and the zero −2 is called a zero of multiplicity 2. The polynomial g(x) = (x + 1)³(x – 2) has zeros −1 and 2 with multiplicities 3 and 1 respectively.

Slide6

Complete Factored Form

has

n

real zeros

c

1

,

c

2

,

c

3

, …,

c

n

, where a distinct zero is listed as many times as its multiplicity. Then

f

(

x

) can be written in

complete factored form

as

f

(

x

) =

a

n

(

x

−

c

1

)(

x

−

c

2

)(

x

−

c

3

) ···(

x

−

c

n

).

Slide7

Example: Finding a complete factorization

Write the complete factorization for the polynomial

f

(

x

) =

7

x

³

−

21

x

²

−

7

x

+ 21 with given zeros

−

1

, 1 and 3.

Solution

Leading

coefficient is

7

.

Zeros

are

−

1

, 1 and 3.

The

complete

factorization:

f

(

x

) =

7

(

x

+ 1)(

x

−

1)(

x

−

3

).

Slide8

Example: Factoring a polynomial graphically

Use the graph of f to factor f(x) = 2x³ − 4x² − 10x + 12.

Solution

Leading coefficient is

2. Zeros

are

−

2, 1, and

3. The

complete

factorization:

f

(

x

) = 2(

x

+ 2)(

x

−

1)(

x

−

3).

Slide9

Example: Factoring a polynomial symbolically

The polynomial f(x) = 2x³ − 2x² − 34x − 30 has a zero of −1. Express f(x) in complete factored form.SolutionIf −1 is a zero, by the factor theorem (x + 1) is a factor. Use synthetic division.

Slide10

Rational Zeros Test (1 of 2)

Slide11

Rational Zeros Test (2 of 2)

Slide12

Example: Finding rational zeros of a polynomial (1 of 3)

Find all rational zeros of f(x) = 6x³ − 5x² − 7x + 4 and factor f(x).Solutionp is a factor of 4, q is a factor of 6

Any rational zero must occur in the list

Slide13

Example: Finding rational zeros of a polynomial (2 of 3)

Evaluate f(x) at each value in the list.

Slide14

Example: Finding rational zeros of a polynomial (3 of 3)

Slide15

Descartes’ Rule of Signs (1 of 2)

Let

P

(

x

) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of

x

.

a

. The

number of positive real zeros either equals the number of variations in sign occurring in the coefficients of

P

(

x

) or is less than the number of variations by a positive even integer

.

Slide16

Descartes’ Rule of Signs (2 of 2)

Let

P

(

x

) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of

x

.

b

.

The number of negative real zeros either equals the number of variations in sign occurring in the coefficients of

P

(

−

x

) or is less than the number of variations by a positive even integer

.

Slide17

Example: Applying Descartes’ rule of signs (1 of 2)

Solution

P(x) has three variations in sign

Thus,

P

(

x

) has 3, or 3

−

2 = 1 positive real zeros.

Slide18

Example: Applying Descartes’ rule of signs (2 of 2)

For negative zeros, consider the variations in sign for P (−x)

P

(

−

x

)

has one variation in sign,

P

(

x

) has only 1 negative real zero.

Slide19

Polynomial Equations

Factoring can be used to solve polynomial equations with degree greater than 2

.

Slide20

Example: Solving a polynomial equation (1 of 2)

Solution

Slide21

Example: Solving a polynomial equation (2 of 2)

Slide22

Example: Solving a polynomial equation (3 of 3)

Slide23

Example: Finding a solution graphically

Since there is only one

x-intercept the equation has one real solution: x  2.65

Slide24

Intermediate Value Theorem

Let (

x

1

,

y

1

) and (

x

2

,

y

2

) with

y

1

≠

y

2

and

x

1

<

x

2

, be two points on the graph of a continuous function

f

. Then, on the interval

x

1

≤

x

≤

x

2

,

f

assumes every value between

y

1

and

y

2

at least once

.

Slide25

Applications: Intermediate Value Theorem

There are many examples of the intermediate value theorem. Physical motion is usually considered to be continuous. Suppose at one time a car is traveling at 20 miles per hour and at another time it is traveling at 40 miles per hour. It is logical to assume that the car traveled 30 miles per hour at least once between these times. In fact, by the intermediate value theorem, the car must have assumed all speeds between 20 and 40 miles per hour at least once

.