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Chapter 6

Polynomials and Polynomial

Functions

Lesson 6-1

Polynomial Functions

Polynomials

A polynomial is a monomial or the sum of monomials.

1

1 1 0( ) ....n n

n nP x a x a x a x a

The exponent of the variable in a term determines the

degree of the term.

A polynomial shown in descending order by degree is in

standard form.

3 2( ) 2 5 2 5P x x x x

Polynomials

3 25( ) 522P xx xx

Leading

Coefficient

Cubic

TermQuadratic

Term

Linear

Term Constant

Degree of a Polynomial

Degree Name

Using

Degree

Example Number

of

Term

Name Using

Number of

Terms

0 Constant 1 Monomial

1 Linear 2 Binomial

2 Quadratic 2 Binomial

3 Cubic 3 Trinomial

4 Quartic 2 Binomial

5 Quintic 4 Polynomial of

4 terms

3x

23 2x

6

3 22 5 2x x x

4 23x x

5 22 3 4x x x

Example 1 – Page 303, #2

Write each polynomial in standard form. Then classify

it by degree and by number terms.

5 3x

3 5x linear; 2 terms; binomial

Example 1 – Page 303, #12

Write each polynomial in standard form. Then classify

it by degree and by number terms.

2 4 22x x x

4 23x x quartic; 2 terms; binomial

Example – Page 304, #34

Simplify. Classify each result by number of terms.

3 38 7 6d d

3 38 7 6d d

39 13d

2 terms; Binomial

Example – Page 304, #42

Simplify. Classify each result by number of terms.

3 4 312 5 23 4 31 9x x x x

4 33 4 312 5 23 1 9x x xx

4 34 3 5 54x x x

4 terms; Polynomial of 4 terms

Example – Page 304, #46

Find each product. Classify the result by number of terms.

2 4 1x x x

22 4 1x x

3 28 2x x

2 terms; binomial

Example – Page 304, #58

Find each product. Classify the result by number of terms.

s t s t s t s t

2 2 2 2s t s t

4 2 2 42s s t t

3 terms, trinomial

Lesson 6-2, Part 1

Polynomial and Linear Functions

Example 1 – Page 311, #6

Write each expression as a polynomial in standard form.

1 1x x x

2 1x x

3x x

Example 2 – Page 311, #8

Write each polynomial in factored form.

3 29 6 3x x x

23 3 2 1xx x

3 3 1 1x x x

Example 4 – Page 311, #18

Find the zeros of each function.

5 8y x x x

0x 5 0x 8 0x

5x 8x

The zeros of the function are 0, -5, and 8

Lesson 6-2, Part 2

Polynomial and Linear Functions

Example 5 – Page 311, #22

Write a polynomial function in standard form with the given

zeros.

2,0,1x

22 0x

x

0

0 0x

x

1

1 0x

x

( ) 2 1

2 1

f x x x x

x x x

2 2x x x 3 2 2x x x

Multiplicity

A repeated zero is called a multiple zero. A multiple zero

has a multiplicity equal to the number of times the zero

occurs.

2( ) ( 2)( 1)( 1)f x x x x

2x 1x

Since x = –1 occurs 3 times it has a multiplicity of 3

Example 6 – Page 311, #30

Find the zeros of each function. State the multiplicity of

multiple zeros.

3( 1)y x x

0x 1 01

xx

Multiplicity of 3

Example 6 – Page 311, #32

Find the zeros of each function. State the multiplicity of

multiple zeros.

33 3y x x

23 1x x

3 1 1x x x

3 00

xx

1 01

xx

1 0

1x

x

Lesson 6-3, Part 1

Dividing Polynomials

Example 1 – Page 315, #6

Divide using long divisions.

29 21 20 1x x x

29 21 201x x x

9x

29 9x x

12 20x

12

12 12x

32

R –32

Example 1 – Page 315, #8

Divide using long divisions.

3 13 12 4x x x

3 24 0 13 12x xx x

2x

3 24x x24 13x x

4x

24 16x x

3 12x

3

3 12x

0

Lesson 6-3, Part 2

Dividing Polynomials

Synthetic Division

Allows you divide by linear factor. You omit all variables

and exponents and reverse the sign of the divisor, so you

can add throughout the process.

3 22 2 5 3x x x x

32 2 5 1

Example 3 – Page 318, #14

Divide using synthetic division

3 24 6 4 2x x x x

2 1 4 6 4

1

2

2

4

2

4

0

2 2 2x x

Example 3 – Page 318, #22

Divide using synthetic division

26 8 2 1x x x

1 6 8 2

6

6

2

2

4

6 2, 4x R

Example 4 – Page 318, #24

Use synthetic division to completely factor

each polynomial function.

3 24 9 36; 3y x x x x

3 1 4 9 36

1

3

7

21

12

36

0

2( 3)( 7 12)y x x x

( 3)( 4)( 3)y x x x

2 7 12x x

Remainder Theorem

If a polynomial P(x) of degree n ≥ 1 is divided by (x – a),

where a is a constant, then the remainder is P(a).

Example 5 – Page 318, #29

Use synthetic division and Remainder Theorem to find P(a)

3 2( ) 7 4 ; 2P x x x x a

2

2

1 5

10

6

1 7 4 0

12

12

3 2( 2) 2 7( 2) 4 2

12

( )P

Lesson 6-4

Solving Polynomial Equations

Sum or difference of Cubes

3 3 2 2

3 3 2 2

a b a b a ab b

a b a b a ab b

Example 3 – Page 324, #14

Factor each expression

3125 27x

3 3

5 3x

3 3 2 2a b a b a ab b

25 3 25 15 9x x x

53

a xb

Example 4– Page 324, #16

Solve each equation

3 64 0x

3 3

4x

3 3 2 2a b a b a ab b

24 4 16x x x

4a xb

2( ) 4

2

b b acx

a

4 16 4(1)(16)

2(1)

4 48 4 16 3

2 2

4 4 32 2 3

2

x

i

ii

4 04

xx

2 2 3x i

Example 5 – Page 324, #22

Factor each expression

4 28 20x x

2 210 2x x

Example 6 – Page 324, #32

Solve each equation

4 28 15 0x x

2 23 5 0x x

2 3 0x 2 5 0x

2 3

3

x

x i

2 5

5

x

x i

Lesson 6-5, Part 1

Theorems About Roots of

Polynomial Equations

Rational Root Theorem

Both the constant term and the leading coefficient

of polynomial can help identify the rational roots of a

polynomial equation with integer coefficients.

Possible rational roots of the equation have the form:

Factor of the constant

Factor of the Leading Coefficient

Example 1 – Page 333, #2

Use the Rational Root Theorem to list all possible rational

roots for each polynomial equation. Then find any

actual rational roots.3 24 6 0x x x

6

1

1,2,3,61,2,3,6

1

Factor of

3 2( ) 4 6P x x x x 3 2(1) 1 4(1) 1 6 0P

3 2( 2) 2 4( 2) ( 2) 6 0P

3 2

( 3) 3 4 3 3 6 0P

Actual solutions are

1, 2, 3

Example 1 – Page 333, #4

Use the Rational Root Theorem to list all possible rational

roots for each polynomial equation. Then find any

actual rational roots.

3 22 9 11 8 0x x x

8

2

1 2 4 8, , ,

2 2 2

1,2,4,8 1, 1,2,4,8,

, 21,2,

1 24 8

2,

Factor of

No rational roots

Example 2 – Page 333, #8

Find the roots of each polynomial equation.

3 25 7 35 0x x x

Step 1 – List all possible rational roots.

35

1

1,5,7,351,5,7,35

1

Example 2 – Page 333, #8

3 2( ) 5 7 35P x x x x

Step 2 – Use the Remainder Root Theorem to find a root

1,5,7,35

3 2

(5) 5 5 5 7 5 35 0P

Step 3 – Use synthetic division with the root you

found in step 2.

5 1 5 7 35

1 0 7

2

2

5 0 7

5 7

x x x

x x

3 25 7 35x x x

5 0 35

0

Example 2 – Page 333, #8

Step 4 – Find the roots

25 7 0x x

5 0

5

x

x

2

2

7 0

7

7

x

x

x i

Lesson 6-5, Part 2

Theorems About Roots of

Polynomial Equations

Irrational Root Theorem

Use the quadratic formula to solve the following quadratic

2 4 1 0

2 5 2 5

x x

and

Number pairs of the form and are called

conjugates.

a b a b

Imaginary Root Theorem

Number pairs of the form and are called

complex conjugates.

a bi a bi

Example 3 – Page 333, #14, 18

A polynomial equation with rational coefficients has the

given roots. Find two additional roots.

4 6 3and

4 6, 3

4 3 7i and i

4 ,3 7i i

Example 5 – Page 333, #20

Find a third-degree polynomial equation with rational

coefficients that has the given number of roots.

5 1and i

15, 1 , x ix x i

5 1 1 0x x i x i

2

2

1

x x xi

x i

xi i i

x 1 i

x

1

i

25 2 2 0x x x 3 2

2

2 2

5 10 10

x x x

x x

2x 2x 2

x

53 23 8 10x x x

5 0x 1 0x i 1 0x i

Lesson 6-6

The Fundamental Theorem

of Algebra

Complex Number System

Real Numbers Imaginary

Numbers

Rational Numbers Irrational Numbers

Rational Numbers: 8

5, 0, , 93

Irrational Numbers: 3, 5

Imaginary Numbers: 4 , 3 2 , 2 2i i i

The Fundamental Theorem

of Algebra

Any polynomial of a degree n ≥ 1 has at least

one complex root.

A nth degree polynomial equation has exactly

n roots.

For the equation state the

number of complex roots, the possible number of real

roots and the possible rational roots.

5 4 22 4 4 5 0,x x x

Step 1: Identify the number of complex roots

5th degree polynomial → 5 complex roots

Example 1 – Page 337, #4

Step 2: Identify the possible number of real roots

Real Roots Imaginary Roots

5 0 = 5 roots

3 2 = 5 roots

1 4 = 5 roots

Possible number of real roots: 1, 3, 5

Example 1 - Page 337, #4

(Rational) Irrational

Step 3: Identify the possible number of rational roots

5 4 24 4 02 5x x x

:

: 2

5

factor

f ors

s

act

(1, 5)

(1, 2)

1 51, ,5,

2 2

1 1 5 5, , ,

1 2 1 2

Example 1 - Page 337, #4

Solving Polynomial Equations

If possible factor out any common terms.

If the polynomial is a quadratic (2nd Degree) Factoring

Square Root

Quadratic Formula

If the polynomial is cubic (3rd Degree) Sum and difference of a cube

Synthetic Division

If the polynomial is a quartic (4th Degree) Factoring using quadratic form

Synthetic Division

If the polynomial is a 5th degree or higher Synthetic Division

Example 2 – Page 337, #12Find all the zeros of each function.

3 22 3 6y x x x

Step 1 – Identify possible rational roots.

: 6 (1, 2, 3, 6)

: 1 (1)

factor

factor

Step 2 – Use the Remainder Root Theorem to find a root.

3 22 2(2) 3(2) 6 0y2x

Example 2 – Page 337, #12

3 22 3 6y x x x

Step 3 – Use Synthetic division using the root from step 2.

2 1 2 3 6 2 1 2 3 6

2 0 6

1 0 3 0

2( 2)( 3) 0x x

Example 2 – Page 337, #12

2 3x

Step 4 – Solve

2( 2)( 3) 0x x

2 3 0x

3x

2 3x

2 0x

2x

The zeros are: 2, 3

Example 2 – Page 337, #14

Find all the zeros of each function.

4 2( ) 3 4f x x x

Step 1 – Factor.

2 21 4 0x x

Example 2 – Page 337, #14

Step 2 – Solve.

2 21 4 0x x

2 1 0x

2 1x

2 1x

1x i

2 4 0x

2 4x

2 4x

2x

The zeros are: , 2i

Lessons

6-1 through 6-6

Graphing Polynomial Functions

Steps to Graphing

Polynomial Equations

Step 1 – Find the end behavior of the graph.

The end behavior is the extreme right or left of

the graph.

The degree of the leading term is odd:

The leading coefficient is positive: the

graph falls to the left and rises to the right.

The leading coefficient is negative: the

graph rises to left and falls to the right.

Steps to Graphing

Polynomial Equations

The degree of the leading term is even:

The leading coefficient is positive: the

graph rises to the left and rises to the right

The leading coefficient is negative: the

graph falls to left and falls to the right

Steps to Graphing

Polynomial Equations

Step 2 – Find the x-intercepts of the graph. This is

where the graph crosses or touches the x-axis.

Find the zeros of the polynomial by:

Set f(x) = 0 and factor

Synthetic Division

Find multiplicity by:

Even multiplicity the graph touches the x-intercept

at this point.

Odd multiplicity the graph crosses the x-intercept

at this point.

Steps to Graphing

Polynomial Equations

Find the y-intercept of the graph. This is

where the graph crosses the y-axis.

Let x = 0 and solve.

Steps to Graphing

Polynomial Equations

Step 4 – Find the maximum number of

turning points. This is where the graph

changes directions.

Maximum number of turning points = Degree of

the leading term – 1

Step 5 – Find additional points if needed.

Example 1 - Graph

3 22 2f x x x x

Step 1 – Find the end behavior of the graph

Leading Term: 3x

Odd degree and positive

Falls to the left and rises to the right:

Example 1 – Graph

1 1 2 1 2

1 3 2

1 3 2 0

Possible rational roots:

Step 2 – Find the x-intercepts of the graph

(1, 2)

1 1 2 1 2

21 3 2 0x x x

3 22 2f x x x x

Example 1 – Graph

1 2 1x x x

:1 :1 :1mult mult mult

2 31 2 0x xx

( 1)( 2)( 1) 0x x x

1 0 2 0 1 0x x x

The graph crosses the x-intercepts at: ( 1, 2)

Example 1 – Graph

Let 0x

Step 3 – Find the y-intercept of the graph

3 20 2(0) 0 2

2

y

y

The y-intercept is: 2

3 22 2f x x x x

Example 1 – Graph

Leading term: 3x

Turning points is: 2

Step 4 – Find the maximum number of turning points.

Maximum number of turning points: 3 1 2

3 22 2f x x x x

Example 1 – Graph

( 1, 2)

2

x-intercepts at: crosses

y-intercept is:

Turning points is:

End behavior:

2

3 22 2f x x x x

Example 1 – Graph

( 1, 2)

2

x-intercepts at: crosses

y-intercept is:

Turning points is:

End behavior:

2

3 22 2f x x x x

Example 2 – Graph

4 216f x x x

Step 1 – Find the end behavior of the graph

Leading Term: 4x

Even degree and negative

Falls to the left and falls to the right:

Example 2 – Graph

0 4 4x x x

: 2 :1 :1mult mult mult

4 2( 16 ) 0x x

2 2 16( ) 0xx

2 0 4 0 4 0x x x

The graph crosses x-intercepts at: and touches at: ( 4)

Step 2 – Find the x-intercepts of the graph

2( 4)( 4) 0x x x

0

4 216f x x x

Example 2 – Graph

Let 0x

Step 3 – Find the y-intercept of the graph

4 20 16(0)

0

y

y

The y-intercept is: 0

4 216f x x x

Example 2 – Graph

Leading term: 4x

Turning points is: 3

Step 4 – Find the maximum number of turning points.

Maximum number of turning points: 4 1 3

4 216f x x x

Example 2 – Graph

4 216f x x x

( 4)

0

x-intercepts at: crosses

y-intercept is:

Turning points is: 3

End behavior:

x-intercepts at: touches (0)

Example 2 – Graph

4 216f x x x

( 4)

0

x-intercepts at: crosses

y-intercept is:

Turning points is: 3

End behavior:

x-intercepts at: touches (0)

Homework

Graph the function.

3 21. ( ) 4 4f x x x x

4 3 22. ( ) 6 9f x x x x

5 33. ( ) 6 9f x x x x

4 34. ( ) 2 2f x x x

Answers1. 2.

3. 4.