Copyright © 2011 Pearson Education, Inc. Slide 3.6-1
Copyright © 2011 Pearson Education, Inc. Slide 3.6-2
Chapter 3: Polynomial Functions
3.1 Complex Numbers3.2 Quadratic Functions and Graphs3.3 Quadratic Equations and Inequalities3.4 Further Applications of Quadratic Functions and Models3.5 Higher-Degree Polynomial Functions and Graphs3.6 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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-3
3.6 Topics in the Theory of Polynomial Functions (I)
The Intermediate Value Theorem
If P(x) defines a polynomial function with only real coefficients, and if, for real numbers a and b, the values P(a) and P(b) are opposite in sign, then there exists at least one real zero between a and b.
Copyright © 2011 Pearson Education, Inc. Slide 3.6-4
3.6 Applying the Intermediate Value Theorem
Example Show that the polynomial function defined by has a real zero between 2 and 3.
Analytic SolutionEvaluate 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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-5
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.
Copyright © 2011 Pearson Education, Inc. Slide 3.6-6
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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-7
3.6 Division of Polynomials
5.down bring andSubtract
.9step,next In the
525)3(9279
2993
9352033
2
2
23
2
23
9 2
xxxxx
xxxx
xxxxxx
xxx
Copyright © 2011 Pearson Education, Inc. Slide 3.6-8
3.6 Division of Polynomials
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.
Copyright © 2011 Pearson Education, Inc. Slide 3.6-9
3.6 Division of Polynomials
We can rewrite the solution as
Divisor Dividend
.
3802593
3523
quotient theofpart
FractionalPolynomialQuotient
23
xxx
xxx
Divisor Remainder
Division of a Polynomial by x – k1. If the degree n polynomial P(x) is divided by x – k,
then 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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-10
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
2993
520332
2
2
23
23
xx
xxxxxx
xxxxxx
2593
807525
525279
2993
520331
Copyright © 2011 Pearson Education, Inc. Slide 3.6-11
3.6 Synthetic Division
• This abbreviated form of long division is called synthetic division.
3802593
80259375279
52033
2
xxx
Copyright © 2011 Pearson Education, Inc. Slide 3.6-12
3.6 Using Synthetic Division
Example Use synthetic division to divideby 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.
Copyright © 2011 Pearson Education, Inc. Slide 3.6-13
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
xxx
xxx
.82865 23 xxx
Copyright © 2011 Pearson Education, Inc. Slide 3.6-14
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).
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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-15
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.
Copyright © 2011 Pearson Education, Inc. Slide 3.6-16
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)
Copyright © 2011 Pearson Education, Inc. Slide 3.6-17
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 TheoremThe polynomial P(x) has a factor x – k if and only if P(k) = 0.
Copyright © 2011 Pearson Education, Inc. Slide 3.6-18
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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-19
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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-20
3.6 Relationships Among x-Intercepts, Zeros, and Solutions
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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-21
3.6 Relationships Among x-Intercepts, Zeros, and Solutions
(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
Copyright © 2011 Pearson Education, Inc. Slide 3.6-22
3.6 Division of Any Two Polynomials
Division Algorithm for PolynomialsLet P(x) and is D(x) be two polynomials, with the degree of D(x) greater than zero and less than the degree of P(x).Then there exist unique polynomials Q(x) and R(x) such that
where either R(x) = 0 or the degree of R(x) is less than thedegree of D(x).
( ) ( )( ) ,( ) ( )
P x R xQ xD x D x
Copyright © 2011 Pearson Education, Inc. Slide 3.6-23
3.6 Dividing Polynomials
Divide
Solution
The quotient is 3x – 2 with remainder – 4. This result can also be written
as
26 5 10 .2 3
x xx
2
2
3 22 3 6 5 10
6 94 104 6
4
xx x x
x xxx
43 2 .2 3
xx
Copyright © 2011 Pearson Education, Inc. Slide 3.6-24
3.6 Dividing Polynomials
Divide
Solution
The quotient is 5x – 4 with remainder 2x + 2. This result can also be written
as
3 2 25 4 7 2 1 .x x x x
2 3 2
3 2
2
2
5 40 1 5 4 7 2
5 0 54 2 24 0 4
2 2
xx x x x x
x x xx xx x
x
2
2 25 4 .1
xxx