chapter 3: polynomial functions


Chapter 3: Polynomial Functions

3.1 Complex Numbers

3.2 Quadratic Functions and Graphs

3.3 Quadratic Equations and Inequalities

3.4 Further Applications of Quadratic Functions and Models

3.5 Higher Degree Polynomial Functions and Graphs

3.6 Topics in the Theory of Polynomial Functions (I)

3.7 Topics in the Theory of Polynomial Functions (II)

3.8 Polynomial Equations and Inequalities; Further Applications and Models


3.6 Topics in the Theory of Polynomial Functions (I)

The Intermediate Value Theorem

If P(x) defines a polynomial function, and if P(a) P(b), then for a x b, P(x) takes every value between P(a) and P(b) at least once.


3.6 Applying the Intermediate Value Theorem

Example Show that the polynomial function defined by

has a real zero between 2 and 3.

Analytic Solution

Evaluate P(2) and P(3).

Since P(2) = –1 and P(3) = 7 differ in sign, the intermediate value theorem assures us that there is a real zero between 2and 3.

12)( 23 xxxxP

713)3(23)3(112)2(22)2(

23

23

PP


3.6 Applying the Intermediate Value Theorem

Graphing Calculator Solution

We see that the zero lies between 2.246 and 2.247 since there is a sign change in the function values.

Caution If P(a) and P(b) do not differ in sign, it does NOT imply that there is no zero between a and b.


3.6 Division of Polynomials

Example Divide the polynomial 3x3 – 2x + 5 by x – 3.

.2down Bring29

93

352033

Subtract.9

)3(393

352033

.3 Start with

0 termMissing52033

2

23

2

23

2

223

2

23

23

223

3

x-xx

xx

xxxxx

x

xxxx

xxxxx

x

xxxxx

xx



5.down bring andSubtract

.9step,next In the

525)3(9279

2993

9352033

2

2

23

2

23

9 2

xxxxx

xxxx

xxxxxx

xxx



Subtract. 80)3(257525

525279

2993

259352033

.25 Finally,

2

2

23

2

23

25

xxxxxxx

xx

xxxxxx

xx

The quotient is 3x2 + 9x +25 with a remainder of 80.



We can rewrite the solution as

Divisor Dividend

.

380

25933

523

quotient theofpart

FractionalPolynomialQuotient

23

xxx

xxx

Divisor Remainder

Division of a Polynomial by x – k

1. If the degree n polynomial P(x) is divided by x – k, the quotient polynomial, Q(x), has degree n – 1.

2. The remainder R is a constant (and may be 0). The complete quotient for may be written as

kxxP

)(

.)()(

kxR

xQkxxP


3.6 Synthetic Division

• The condensed version of previous example:

Dropping the variables, we see the repetition of numbers. We condense long division of a polynomial by x – k by replacing subtraction with addition and changing the sign of –3 to 3.

2593

807525

525279

29

93

52033

2

2

2

23

23

xx

xxxx

xx

xx

xxxx2593

807525

525279

2993

520331


3.6 Synthetic Division

• This abbreviated form of long division is called synthetic division.

380

2593

80259375279

52033

2

xxx


3.6 Using Synthetic Division

Example Use synthetic division to divide

by x + 2.

Solution x + 2 = x – (–2), so k = –2.

5

828652

82865 23 xxx

16510

828652

Bring down the 5.

Multiply –2 by 5 to get –10 and add it to –6.


3.6 Using Synthetic Division

• Notice that x + 2 is a factor of

41653210

828652

Multiply –2 by –16 to get 32 and add it to –28.

0416583210828652

Multiply –2 by 4 to get –8 and add it to 8.

41652

82865 223

xx

xxxx

.82865 23 xxx


3.6 The Remainder Theorem

Example Use the remainder theorem and synthetic division

to find P(–2) if

Solution Use synthetic division to find the remainder when

P(x) is divided by x – (–2).

The Remainder Theorem

If a polynomial P(x) is divided by x – k, the remainder is equal to P(k).

.543)( 24 xxxxP

121214242543012

.1)2(

theorem,remainder By the

.1 isremainder The

P


3.6 k is Zero of a Polynomial Function if P(k) = 0

ExampleDecide whether the given number is a zero of P.

Analytic Solution

(a)

(b)

1094)(;2 (a) 23 xxxxP

xxxxP23

23

)(;2 (b) 23

05211042109412

The remainder is zero, sox = 2 is a zero of P.

1941983012

219

23

23

23

The remainder is not zero, sox = –2 is not a zero of P.


3.6 k is Zero of a Polynomial Function if P(k) = 0

Graphical Solution

Y1 = P(x) in part (a)

Y2 = P(x) in part (b)

Y1(2) = P(2) in part (a)

Y2 (-2) = P(-2) in part (b)


3.6 The Factor Theorem

• From the previous example, part (a), we have

indicating that x – 2 is a factor of P(x).

),52)(2()(522)( 22

xxxxPxx

xxP

The Factor Theorem

The polynomial x – k is a factor of the polynomial P(x) if and only if P(k) = 0.


3.6 Example using the Factor Theorem

Determine whether the second polynomial is a factor of the first.

Solution Use synthetic division with k = –2.

Since the remainder is 0, x + 2 is a factor of P(x), where

2;3248244)( 23 xxxxxP

0161643232832482442

).16164)(2(3248244)(

2

23

xxxxxxxP


3.6 Relationships Among x-Intercepts, Zeros, and Solutions

Example Consider the polynomial function

(a) Show by synthetic division that are zeros of P, and write P(x) in factored form.

(b) Graph P in a suitable viewing window and locate the x- intercepts.

(a) Solve the polynomial equation

.652)( 23 xxxxP

1 and ,,2 23

.0652 23 xxx



Solution(a)

031262461522

0)2( P

02233312

23

0)( 23 P

022221

0)1( P

).1)(23)(2(2)( So factor.constant theis 2 This xxxxP



(b) The calculator will determine the x-intercepts: –2, –1.5, and 1.

(c) Because the zeros of P are the solutions of P(x) = 0, the solution set is }.1,,2{ 2

3

chapter 3: polynomial functions

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