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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


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


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


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


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


Write each expression as a polynomial in standard form.

1 1x x x

2 1x x

3x x


Write each polynomial in factored form.

3 29 6 3x x x

23 3 2 1xx x

3 3 1 1x x x


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


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


Find the zeros of each function. State the multiplicity of

multiple zeros.

3( 1)y x x

0x 1 01

xx

Multiplicity of 3


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


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


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


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


Divide using synthetic division

26 8 2 1x x x

1 6 8 2

6

6

2

2

4

6 2, 4x R


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).


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


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


Factor each expression

4 28 20x x

2 210 2x x


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


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


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


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


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


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


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


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


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


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


2 3x

Step 4 – Solve

2( 2)( 3) 0x x

2 3 0x

3x

2 3x

2 0x

2x

The zeros are: 2, 3


Find all the zeros of each function.

4 2( ) 3 4f x x x

Step 1 – Factor.

2 21 4 0x x


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


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


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


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


Find the y-intercept of the graph. This is

where the graph crosses the y-axis.

Let x = 0 and solve.

Steps to Graphing


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 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


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:


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.

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