Week 4. Due for this week… Homework 4 (on MyMathLab – via the Materials Link) Monday night at 6pm. Prepare for the final (available tonight 10pm

Week 4

Due for this week…

Homework 4 (on MyMathLab – via the Materials Link) Monday night at 6pm.

Prepare for the final (available tonight 10pm to Saturday Aug 20th 11:59pm)

Do the MyMathLab Self-Check for week 4. Learning team presentations week 5.

Slide 2Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Final Exam logistics

Here is what I've found out about the final exam in MyMathLab (running from the end of class this week (week 4 at 10pm) to Saturday night 8/20/2011 at 11:59pm (the first Saturday after the last day of class).

.


Final Exam logistics There will be 50 questions. You have only one attempt to complete the exam. Once you start the exam, it must be completed in that sitting. (Don't start until you have

time to complete it that day or evening.) You may skip and get back to a question BUT return to it before you hit submit.

You must be in the same session to return to a question. There is no time limit to the exam (except for 11:59pm Saturday night after the last class). You will not have the following help that exists in homework:

Online sections of the textbook Animated help Step-by-step instructions Video explanations Links to similar exercises

You will be logged out of the exam automatically after 3 hours of inactivity. Your session will end.

IMPORTANT! You will also be logged out of the exam if you use your back button on your browser. You session will end.


Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Rules for Exponents

Review of Bases and Exponents

Zero Exponents

The Product Rule

Power Rules

5.1


Review of Bases and Exponents

The expression 53 is an exponential expression with base 5 and exponent 3.

Its value is 5 5 5 = 125.

bn

Base

Exponent

times

...n

b b b b


EXAMPLE Evaluating exponential expressions

Evaluate each expression. a. b. c.

Solutiona. b. c.

242

8

43 4( 3)

242

8

2 factors

28

4 4

162

8

2 2 4

434 factors

3 3 3 3( )

81

4 factors

( 3) ( 3) ( 3) ( 3)

81

4( 3)

Try some of Q: 11-16, 19-26


Zero Exponents


EXAMPLE Evaluating exponential expressions

Evaluate each expression. Assume that all variables represent nonzero numbers. a. b. c.

Solution

a. b. c.

08

02

43

03 7

2

x y

z

080

24

3

03 7

2

x y

z

1

4(1) 4

1

Try some of Q: 17-18


The Product Rule


EXAMPLE Using the product rule

Multiply and simplify. a. b. c.

Solution

a. b. c.

2 43 32 73 6x x 2 2(3 4 )x x x

2 43 3 2 73 6x x 2 2(3 4 )x x x2 43 3

2 73 6 x x

2 43 63

729

2 718x

918x

2 2 23 4x x x x

3 43 4x x



Exponent Rules


EXAMPLE Raising a power to a power

Simplify the expression. a. b.

Solution

a. b.

323 52x

323 52x

2 33 2 5x

6310x



Exponent Rules


EXAMPLE Raising a product to a power

Simplify the expression. a. b. c.

Solution

a. b. c.

3(2 )a 2 3( 3 )x 3 4 2( 2 )h

3(2 )a 2 3( 3 )x 3 4 2( 2 )h3(2 )a

3 32 a

38a

2 3( 3 )x

3 2 3( 3) ( )x

627x

4 2( 8 )h 2 4 2( 8) ( )h

864h



Exponent Rules


EXAMPLE Raising a quotient to a power


Solution

a. b. c.

33

4

7a

b

3

4

x y

33

4

7a

b

3

3

( )

4

x y

3

3

3 27

4 64

7

7

a

b

3

4

x y

3( )

64

x y



EXAMPLE Combining rules for exponents


Solution

a. b. c.

2 3(3 ) (4 )a a

32 4a b

d

2 2 3 4 3(3 ) ( 5 )a b a b

2 32 3 3 32 2 43 ( ) ( 5) ( ) ( )a b a b 2 2 3 33 4a a 2 3 4 3

3

( ) ( )a b

d

2 3(3 ) (4 )a a

32 4a b

d

2 2 3 4 3(3 ) ( 5 )a b a b

6 12

3

a b

d

4 2 9 129 ( 125)a b a b 4 9 2 129( 125)a a b b

13 141125a b

2 39 64a a 5576a



Addition and Subtraction of Polynomials

Monomials and Polynomials

Addition of Polynomials

Subtraction of Polynomials

Evaluating Polynomial Expressions

5.2

A monomial is a number, a variable, or a product of numbers and variables raised to natural number powers.Examples of monomials: The degree of monomial is the sum of the exponents of the variables. If the monomial has only one variable, its degree is the exponent of that variable.


3 2 9 88, 7 , , 8 , y x x y xy

The number in a monomial is called the coefficient of the monomial.


EXAMPLE

Solution

Identifying properties of polynomials

Determine whether the expression is a polynomial. If it is, state how many terms and variables the polynomial contains and its degree.

a. 9y2 + 7y + 4 b. 7x4 – 2x3y2 + xy – 4y3 c. 2 38

4x

x

a. The expression is a polynomial with three terms and one variable. The term with the highest degree is 9y2, so the polynomial has degree 2. b. The expression is a polynomial with four terms and two variables. The term with the highest degree is 2x3y2, so the polynomial has degree 5.

c. The expression is not a polynomial because it contains division by the polynomial x + 4. Try some of Q: 19-30


EXAMPLE

Solution

Adding like terms

State whether each pair of expressions contains like terms or unlike terms. If they are like terms, then add them.

a. 9x3, −2x3 b. 5mn2, 8m2n

a. The terms have the same variable raised to the same power, so they are like terms and can be combined.

b. The terms have the same variables, but these variables are not raised to the same power. They are therefore unlike terms and cannot be added.

9x3 + (−2x3) = (9 + (−2))x3 = 7x3



EXAMPLE

Solution

Adding polynomials

Add each pair of polynomials by combining like terms. 2 23 4 8 4 5 3x x x x

2 28 3443 5x x x x

2 2 4 8 34 53x x x x

2 23 4 8 4 5 3x x x x

2 4( ) (3 4 )3) (85x x

2 57x x



EXAMPLE

Solution

Adding polynomials vertically

Simplify

Write the polynomial in a vertical format and then add each column of like terms.

2 2 2 27 3 7 2 2 .x xy y x xy y

2

2

2

2

7 3 7

2 2

yxy

yx y

x

x

2 25 2 5xyx y



To subtract two polynomials, we add the first polynomial to the opposite of the second polynomial. To find the opposite of a polynomial, we negate each term.


EXAMPLE

Solution

Subtracting polynomials

Simplify

The opposite of

3 2 3 25 3 6 5 4 8 .w w w w

3 2 3 25 4 8 is 5 4 8w w w w

3 2 3 25 3 6 5 4 8w w w w

3 2(5 5) (3 4) ( 6 8)w w

3 20 7 2w w 27 2w



EXAMPLE

Solution

Subtracting polynomials vertically

Simplify

Write the polynomial in a vertical format and then add the first polynomial and the opposite of the second polynomial.

2 210 4 5 4 2 1 .x x x x

2

2

10 4 5

4 2 1

x

x

x

x

26 6 6x x



Multiplication of Polynomials

Multiplying Monomials

Review of the Distributive Properties

Multiplying Monomials and Polynomials

Multiplying Polynomials

5.3


Multiplying Monomials

A monomial is a number, a variable, or a product of numbers and variables raised to natural number powers. To multiply monomials, we often use the product rule for exponents.


EXAMPLE Multiplying monomials

Multiply. a. b.

Solutiona. b.

4 36 3x x 3 4 2(6 )( )xy x y

4 36 3x x 4 3( 6)(3)x

718x

3 4 2(6 )( )xy x y

4 3 26xx y y1 4 3 26x y

5 56x y

Try some of Q: 7-16


EXAMPLE

Solution

Using distributive properties

Multiply. a. b. c.

a.

3(6 )x 4( 2 )x y (3 5)(7)x

b. 3 36 6( ) 3x x

18 3x

4( ) ( ) ( )( 2 )4 42x y x y

4 8x y

c. 3 5 3( )( ) ( ) ( )757 7x x

21 35x Try some of Q: 17-24


EXAMPLE Multiplying monomials and polynomials

Multiply. a. b.

Solutiona. b. 24 (3 2)xy x y

23 24 4x yxy xy 212 8xx yy xy

3 3( )ab a b

3 3ab a ab b 4 4a b ab

24 (3 2)xy x y 3 3( )ab a b

3 212 8x y xy



Multiplying Polynomials

Monomials, binomials, and trinomials are examples of polynomials.


EXAMPLE

Solution

Multiplying binomials

Multiply ( 2)( 4).x x

2 24 4x x xx

2 2( )( ) ( )( )4 )2 ( )4(x xx x x

2 2 4 8x x x

2 6 8x x




EXAMPLE

Solution

Multiplying binomials

Multiply each binomial.a. b.

a.

(3 1)( 4)x x

(3 1)( 4)x x 3 3 4 1 1 4x x x x

23 12 4x x x 23 11 4x x

2( 2)(3 1)x x

2( 2)(3 1)x x b. 2 23 ( 1) 2 3 2 1x x x x 3 23 6 2x x x



EXAMPLE

Solution

Multiplying polynomials

Multiply each expression. a. b.

a.

24 ( 6 1)x x x

24 4 6 4 1x x x x x

3 24 24 4x x x

2( 2)( 5 2)x x x

b. 2 25 ( 2) 2 2 5 2 2x x x x x x x

3 2 25 2 2 10 4x x x x x

24 ( 6 1)x x x

2( 2)( 5 2)x x x

3 27 8 4x x x


EXAMPLE

Solution

Multiplying polynomials

Multiply

2 23 ( 3 4 ).ab a ab b

2 233 3 43ab aba ab bab 3 2 2 33 9 12a b a b ab

2 23(3 )4a abab b



EXAMPLE

Solution

Multiplying polynomials vertically

Multiply

21 (2 3).x x x

22 3

1

x x

x

22 3x x 3 22 3x x x 3 22 4 3x x x



Special Products

Product of a Sum and Difference

Squaring Binomials

Cubing Binomials

5.4



EXAMPLE

Solution

Finding products of sums and differences

Multiply.

a. (x + 4)(x – 4) b. (3t + 4s)(3t – 4s)

a. We can apply the formula for the product of a sum and difference. (x + 4)(x – 4) = (x)2 − (4)2

= x2 − 16

b. (3t + 4s)(3t – 4s) = (3t)2 – (4s)2

= 9t2 – 16s2

Try some of Q: 9-24


EXAMPLE

Solution

Finding a product

Use the product of a sum and difference to find 31 ∙ 29.

Because 31 = 30 + 1 and 29 = 30 – 1, rewrite and evaluate 31 ∙ 29 as follows.

31 ∙ 29 = (30 + 1)(30 – 1)

= 302 – 12

= 900 – 1

= 899




EXAMPLE

Solution

Squaring a binomial

Multiply.

a. (x + 7)2 b. (4 – 3x)2

a. We can apply the formula for squaring a binomial.

(x + 7)2 = (x)2 + 2(x)(7) + (7)2

b.

= x2 + 14x + 49

(4 – 3x)2 = (4)2 − 2(4)(3x) + (3x)2

= 16 − 24x + 9x2



EXAMPLE

Solution

Cubing a binomial

Multiply (5x – 3)3.

= (5x − 3)(5x − 3)2

= 125x3

(5x – 3)3

= (5x − 3)(25x2 − 30x + 9)

= 125x3 – 225x2 + 135x – 27

– 27 – 150x2 + 45x– 75x2 + 90x



EXAMPLE

Solution

Calculating interest

If a savings account pays x percent annual interest, where x is expressed as a decimal, then after 2 years a sum of money will grow by a factor of (x + 1)2.

a. Multiply the expression.b. Evaluate the expression for x = 0.12 (or 12%), and

interpret the result.

a. (1 + x)2 = 1 + 2x + x2

b. Let x = 0.12 1 + 2(0.12) + (0.12)2 = 1.2544

The sum of money will increase by a factor of 1.2544. For example if $5000 was deposited in the account, the investment would grow to $6272 after 2 years. Try Q: 85


Integer Exponents and the Quotient Rule

Negative Integers as Exponents

The Quotient Rule

Other Rules for Exponents

Scientific Notation

5.5


Negative Integers as Exponents

Simplify each expression.a. b. c.


EXAMPLE Evaluating negative exponents

Solution

a.

b.

c.

521

1

8

4( )a b

52

5

1

2

1

2 2 2 2 2

1

32

1

1

818 8

4( )a b 4

1

( )a b


Evaluate the expression.


EXAMPLE Using the product rule with negative exponents

Solution

4 28 8

4 28 8 4 ( 2)8 28 64


Simplify the expression. Write the answer using positive exponents. a. b.


EXAMPLE Using the rules of exponents

Solution

a.

4 5 6x x x 3 54 3y y

4 5 6x x x 4 ( 5) 6x 5x

b. 3 54 3y y 3 54 3 y y 3 ( 5)12y 212y 2

12

y



Simplify each expression. Write the answer using positive exponents.a. b. c.


EXAMPLE Using the quotient rule

Solution

a.

b.

c.

3

6

10

10

7

3

x

x

2 4

6

24

6

x y

x y

3

6

10

103 610 310

3

1

10

7 3x 4x

2 4

6

24

6

x y

x y

2 4

6

24

6

x y

x y 2 6 4 14x y

1

1000

7

3

x

x

4 34x y3

4

4y

x



Simplify each expression. Write the answer using positive exponents.

a. b. c.


EXAMPLE Working with quotients and negative exponents

Solution

a.

b.

3

1

3

3 6

5 4

2

6

a b

a b

33 27

c.

3 6

5 4

2

6

a b

a b

4 6

5 3

2

6

b b

a a

10

83

b

a

3

1

3

2

3

3

a

2

3

3

a

23

3

a

6

23

a

6

9

a



Important Powers of 10


Number 10-3 10-2 10-1 103 106 109 1012

Value Thousandth Hundredth Tenth Thousand Million Billion Trillion

Write each number in standard form.

a. b.


EXAMPLE Converting scientific notation to standard form

0.0064

Move the decimal point 6 places to the right since the exponent is positive.

3,000,000

Move the decimal point 3 places to the left since the exponent is negative.

63 10 36.4 10



Write each number in scientific notation.

a. 475,000 b. 0.00000325

475000


EXAMPLE Writing a number in scientific notation

0.00000325

63.25 10

Move the decimal point 5 places to the left.

54.75 10

Move the decimal point 6 places to the right.



Division of Polynomials

Division by a Monomial

Division by a Polynomial

5.6


EXAMPLE

Solution

Dividing a polynomial by a monomial

Divide.5 3

2

6 18

6

x x

x

3

2

56 18

6

x x

x

2 2

5 36 8

6 6

1

x x

x x 3 3x x



EXAMPLE Dividing and checking5 4 2

3

16 12 8

4

y y y

y

Check:

5 4 2

3 3 3

16 12 8

4 4 4

y y y

y y y

2 24 3y y

y

3 2 24 4 3y y y

y

3 2 3 3 2

4 4 4 3 4y y y y yy

5 4 216 12 8y y y

Divide the expression and check the result.

5 4 2

3

16 12 8

4

y y y

y

Try some of Q: 9-14


EXAMPLE

Solution

Dividing polynomials

22 1 4 6 8x x x

The quotient is 2x + 4 with remainder −4, which also

can be written as

2x

4x2 – 2x

8x – 8

8x – 4

− 4

+ 4

42 4 .

2 1x

x

Remainder

Quotient + Divisor

Divide and check.24 6 8

2 1

x x

x


EXAMPLE continued

Check:

(Divisor )(Quotient) + Remainder = Dividend

(2x – 1)(2x + 4) + (– 4) = 2x ∙ 2x + 2x ∙ 4 – 1∙ 2x − 1∙ 4 − 4

= 4x2 + 8x – 2x − 4 − 4

= 4x2 + 6x − 8

It checks.



EXAMPLE

Solution

Dividing polynomials having a missing term

Simplify (x3 − 8) ÷ (x − 2).

3 22 0 0 8x x x x

The quotient is

x2

x3 – 2x2

2x2 + 0x 2x2 − 4x

4x − 8

+ 2x + 4

04x − 8

2 2 4.x x Try some of Q: 31-34


EXAMPLE

Solution

Dividing with a quadratic divisor

Divide 3x4 + 2x3 − 11x2 − 2x + 5 by x2 − 2.

2 4 3 20 2 3 2 11 2 5x x x x x x

The quotient is

3x2

3x4 + 0 – 6x2

2x3 − 5x2 − 2x 2x3 + 0 − 4x

−5x2 + 2x + 5

+ 2x − 5

2x – 5 −5x2 + 0 + 10

22

2 53 2 5 .

2

xx x

x


End of week 4

You again have the answers to those problems not assigned

Practice is SOOO important in this course. Work as much as you can with MyMathLab, the

materials in the text, and on my Webpage. Do everything you can scrape time up for, first the

hardest topics then the easiest. You are building a skill like typing, skiing, playing a

game, solving puzzles. NEXT TIME: Team Presentations then MTH 209 for

some the week beyond that.

Documents

Week 4. Due for this week… Homework 4 (on MyMathLab – via the Materials Link) Monday night at 6pm. Prepare for the final (available tonight 10pm