Chapter 6

advertisement

Chapter 6

Polynomials and

Polynomial Functions

6.1 Using Properties of

Exponents

Vocabulary

• Scientific notation – in the form of c

10 n

1 10

• Example: .

 5 

,

Properties of Exponents

• Product of Powers: a m  a n  a

• Power of a Power: m n

( a )

 a mn

• Power of a Product: m  m m

( ab ) a b

• Negative Exponent: a

 m 

1 a m

, a

0

Properties of Exponents

• Zero Exponent: a

0 

1 , a

0

• Quotient of Powers: a m a n

 a , a

0

• Power of a Quotient: ( ) b m  a b m m

, b

0

1.

4 2

( 3 )

Evaluate

2.

5

8 bg

3 

Evaluate

1.

(

2

3

) (

2 )

9 

2.

2 4

( 10 ) ( 10 )

2 

Simplify

1.

( b a

2

3

)

3 

2.

(

 x

2 5 2

12

) x x

1.

xy

2

( xy

)

Simplify

2.

( a b a b

)

3 

Write in Scientific Notation

1. 24,100 =

2. 0.0038 =

A circular component used in the manufacture of a microprocessor has a diameter of 200 mm and a thickness of 0.01 mm. What is it’s volume?

6.2 Evaluating and Graphing

Polynomial Functions

Vocabulary

• Polynomial function

( )

 a x n n  a n

1 x n

1

• Leading coefficient a n

• Constant term

• Degree n a

0

...

a x

 a

1 0

• Standard form – written in descending order of exponents from left to right

1.

Decide if the function is a polynomial. If it is, write in standard form and state its degree, type, and leading coefficient.

f x

2 x

2  x

2

2.

f x

 

.

x

3  x

4 

5

Decide if the function is a polynomial. If it is, write in standard form and state its degree, type, and leading coefficient.

1.

f x

2 x

3 

4 x 

7

2.

f x

8 x

3 

9

Solve Using Direct Substitution f x

2 x

3 

4 x

2 x 7 when x = 3

Solve Using Direct Substitution f x

2 x

3 

5 x

2 

4 x

8 when x = 2

Solve Using Synthetic Substitution f x

2 x

4 

8 x

2 

5 x

7 when x = 3

Solve Using Synthetic Substitution f x

3 x

5  x

4 

5 x

10 when x = -2

Solve Using Synthetic Substitution f x

5 x

3  x

2 

4 x

1 when x = 4

End Behavior

• Behavior of the graph (y-direction) as x approaches positive infinity or negative infinity

( x

 

(

) x

 

)

• “as x approaches positive infinity” ( x

 

)

For a n

End Behavior

0 and even

 

as x

 

and

 

as x

 

For a n

0 and odd

 

as x

 

and

 

as x

 

End Behavior

For a n

0 and even

 

as x

 

and

 

as x

 

For a n

0

 

as x

 

and

 

as x

 

Function f x

 

5 x

3

As x

  f x

2 x

5 

1 f x

2 x

3 x

8 f x

 x

4  x

3

As x

 

Graph f x

 x

3  x

2 

4 x

1

Graph f x

 x

3 

2 x

2 x 3

6.3 Adding, Subtracting, and Multiplying Polynomials

Add the Polynomials

( 5 x

2    

3 x

2 

6 x

1 )

Add the Polynomials

( x

4 

2 x

3  

( 2 x

4 

9 )

Subtract the Polynomials

( 3 x

3 

8 x

2

5 ) ( 5 x

3  x

2 

17 )

Subtract the Polynomials

( 9 x

4 

12 x

3  x

2  

3 x

4 

12 x

3  x )

Multiply the Polynomials

4 x

2  

5 x 2 x

1

Multiply the Polynomials

1.

( 3 x

)( x

2 )

2.

2 5 x

2

+ 3 x 1

Multiply the Polynomials

( x

2 )( x

1 )( x

3 )

Multiply the Polynomials

1.

( 5 x

2 )

2

2.

( 2 x

3 )

3

6.4 Factoring and Solving

Polynomial Equations

Factoring Monomials

3 x

4 

12 x

3

Special Factoring Patterns

• Sum of Two Cubes a

3  b

3 

(

)(

2   2

)

• Example x

3

8 ( x

2 )( x

2 

2 x

4 )

Special Factoring Patterns

• Difference of Two Cubes a

3  b

3 

(

)(

2   2

)

• Example

8 x

3  

( 2 x

)( x

2 

2 x

1 )

x

3 

27

Factor

1.

125

 x

3

Factor

2.

64 x

4 

27 x

Factor by grouping ra

 rb

 sa

 sb

( b ) (

 b )

( r

 s a

 b )

Factor by grouping x

3 

2 x

2 

9 x

18

Factor by grouping

2 2 x y

3 x

2 

4 y

2 

12

1.

Factor the Polynomial

81 x

4 

16

2.

x

4 

4 x

2 

21

3.

4 x

6 

20 x

4 

24 x

2

Solve the Polynomial Equation

1.

2 x

5 

24 x

14 x

3

2.

2 x

5 

18 x

You are building a bin to hold cedar mulch for your garden. The bin will hold 162 cubic feet of mulch. The dimensions of the bin are x ft.

by

5x-6 ft.

by 5x-9 ft.

How tall will the bin be?

6.5 The Remainder and

Factor Theorems

Vocabulary

Polynomial long division - division with a remainder with a lower degree than divisor

Synthetic division – divides polynomial by x-k

Divide Using Long Division

• Divide 297 by 14

Divide Using Long Division

2 x

4 

3 x

3 

5 x

1 by x

2 

2 x

2

Divide Using Long Division x

4 

2 x

2  

5 by x

2  

1

Remainder Theorem

• If a polynomial f(x) is divided by x – k, then the remainder is r = f(k).

Divide Using Synthetic Division x

3 

3 x

2 

7 x

6 by x

4

Divide Using Synthetic Division x

3  x

2 

2 x

8 by x

2

Factor Theorem

• A polynomial f(x) has a factor x – k if and only if f(k) = 0.

Factor the Polynomial f x

3 x

3 

13 x

2 

2 x

8 given that f ( 4 ) 0

Factor the Polynomial f x

3 x

3 

14 x

2 

28 x

24 given that f ( 6 ) 0

Find the Zeros of the Polynomial f x

 x

3 

6 x

2 

3 x

10 One zero is at x

 

5

Find the Zeros of the Polynomial f x

2 x

3 

9 x

2 

32 x

21 One zero is at x

7

6.6 Finding Rational Zeros

Rational Zero Theorem

• Every rational zero of f has the following form: p q

 factor of constant term a

0 factor of leading coefficient a n

List all Possible Rational Zeros f x

 x

4 

2 x

2 

24

List all Possible Rational Zeros

1.

f x

2 x

3 

7 x

2 

7 x

30

2.

f x

6 x

4 

3 x

3 x 10

Find all of the Real Zeros f x

 x

3 

2 x

2 

11 x

12

Find all of the Real Zeros f x

 x

3 

4 x

2 

11 x

30

Find all of the Real Zeros f x

2 x

3 

5 x

2 x 6

Find all of the Real Zeros f x

2 x

3  x

2 

50 x

25

You are building a concrete wheelchair ramp.

The width is three times the height, and the length is 5 feet more that 10 times the height. If

150 cubic feet of concrete is used, what are the dimensions of the ramp?

6.7 Using the Fundamental

Theorem of Algebra

Fundamental Theorem of Algebra

If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one root in the set of complex numbers.

Vocabulary

Repeated solution – occurs more than one time, can be counted twice using the fundamental theorem of algebra.

Polynomial Functions

1. The degree of the polynomial tells you how many solutions it has.

2. Real zeros cross the x-axis.

3.

The graph touches but doesn’t cross the x-axis at repeated zeros.

Find the Zeros x

3 

3 x

2 

16 x

48

0

1.

x

2

Find the Zeros

14 x

49

0

2.

f x

 x

5 

2 x

4 

8 x

2 

13 x

6

Write the Polynomial Function

1. Zeros are at 2, 1, and 4.

2. Zeros are 5, 2i, and -2i

6.8 Analyzing Graphs of

Polynomial Functions

Zeros, Factors, Solutions, Intercepts

• Zero: k is a zero of the polynomial

• Factor: x – k is a factor of the polynomial

• Solution: k is a solution of the polynomial

• X – intercept: if k is a real number, it is an x – intercept of the graph of the polynomial function

Graph the Function f x

1

2

( x

2 )( x

1 )

2

Graph the Function f x

5 ( x

1 )( x

2 )( x

3 )

Graph the Function f x

( x

1

3

) ( x

1 )

Turning Points of Polynomials

• The graph of every polynomial function of degree n has at most n – 1 turning points.

• If a polynomial has n distinct real zeros, then its graph has exactly n – 1 turning points.

Turning Points

Local maximum

Local minimum

6.9 Modeling with Polynomial

Functions

Write the Cubic Function

Write the Cubic Function

Write the Cubic Function

• The graph passes through the given points:

(-1,0), (-1,0), (0,0), and (1,-3)

Write the Cubic Function

• The graph passes through the given points:

(3,0), (2,0), (-1,0), and (1,4)

Properties of Finite Differences

• A polynomial with degree n , has n th order differences of function values for equally spaced x – values that are nonzero and constant

Finding Finite Differences f x

 x

2 

3 x

7

Finding Finite Differences f n

2 n

3  n

2 

2 n

1

Test Review

( b a

2

3

)

3 

(

 x

2 5 2

12

) x x

Simplify xy

2

( xy

)

( a b

 a b

)

3 

4 2

( 3 )

5

8 bg

3 

(

2

3

) (

2 )

9 

2 4

( 10 ) ( 10 )

2 

Solve

Evaluate the Polynomial f x

2 x

4  x

3 

3 x

2 

5 x , when x

 

1

Function f x

 

5 x

3

As x

  f x

2 x

5 

1 f x

2 x

3 x

8 f x

 x

4  x

3

As x

 

Graph f x x 3 x

Complete the Operation

( 9 x

4 

12 x

3  x

2  

3 x

4 

12 x

3  x )

2 5 x

2

+ 3 x 1

( x

2 )( x

1 )( x

3 )

( 2 x

3 )

3

3 x

4 

12 x

3 x

3 

27

125

 x

3

64 x

4 

27 x

Factor

Factor

81 x

4 

16 x

3 

2 x

2 

9 x

18

4 x

6 

20 x

4 

24 x

2

2 2 x y

3 x

2 

4 y

2 

12

Find all Solutions

2 x

5 

18 x

2 x

5 

24 x

14 x

3 f x

2 x

3 

7 x

2 

7 x

30 f x

6 x

4 

3 x

3 x 10

Find all Solutions f x

 x

3 

2 x

2 

11 x

12 f x

 x

3 

4 x

2 

11 x

30 f x

2 x

3 

5 x

2 x 6 f x

2 x

3  x

2 

50 x

25

Factor

12

4 3 x y

20

2 2 x y

24

2 x y

Factoring Tips

1. Always take out GCF first.

2. Two terms – cubic or diff. of two squares

3. Three terms – trinomial (reverse FOIL)

4. Even number of terms – grouping

5. Synthetic always works ( ) q

Download