Download - Week 4 - Biker John
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.
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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).
.
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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.
Slide 4Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Rules for Exponents
Review of Bases and ExponentsZero ExponentsThe Product RulePower Rules
5.1
Slide 6Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 7Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Evaluating exponential expressions
Evaluate each expression. a. b. c.
Solutiona. b. c.
2428
43 4( 3)
2428
2 factors
28
4 4
1628
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
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Zero Exponents
Slide 9Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Evaluating exponential expressions
Evaluate each expression. Assume that all variables represent nonzero numbers.
a. b. c.
Solutiona. b. c.
08024
3
03 7
2x y
z
08024
3
03 7
2x y
z 1
4(1) 4
1
Try some of Q: 17-18
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The Product Rule
Slide 11Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Using the product rule
Multiply and simplify.
a. b. c.
Solutiona. b. c.
2 43 3 2 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
Try some of Q: 27-42
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Exponent Rules
Slide 13Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Raising a power to a power
Simplify the expression.
a. b.
Solutiona. b.
323 52x
323 52x
2 33 2 5x
6310x
Try some of Q: 43-48
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Exponent Rules
Slide 15Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Raising a product to a power
Simplify the expression.
a. b. c.
Solutiona. 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
Try some of Q: 53-62
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Exponent Rules
Slide 17Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Raising a quotient to a power
Simplify the expression.
a. b. c.
Solutiona. b. c.
334
7ab
3
4x y
334
7ab
3
3
( )4
x y
3
3
3 274 64
7
7
ab
3
4x y
3( )64
x y
Try some of Q: 63-74
Slide 18Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Combining rules for exponents
Simplify the expression.
a. b. c.
Solutiona. b. c.
2 3(3 ) (4 )a a32 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 bd
2 3(3 ) (4 )a a32 4a b
d
2 2 3 4 3(3 ) ( 5 )a b a b
6 12
3
a bd
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
Try some of Q: 75-86
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Addition and Subtraction of Polynomials
Monomials and PolynomialsAddition of PolynomialsSubtraction of PolynomialsEvaluating 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.
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3 2 9 88, 7 , , 8 , y x x y xy
The number in a monomial is called the coefficient of the monomial.
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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. 2 384
xx
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
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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
Try some of Q: 31-40
Slide 23Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 41-52
Slide 24Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 72 2
yxyyx y
xx
2 25 2 5xyx y
Try some of Q: 59-56
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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.
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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
Try some of Q: 63-74
Slide 27Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 54 2 1
xx
xx
26 6 6x x
Try some of Q: 75-78
Slide 28Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Writing and evaluating a monomial
Write the monomial that represents the volume of the box having a square bottom as shown. Find the volume of the box if x = 5 inches and y = 3 inches.
The volume is found by multiplying the length, width, and height together. This can be written as x2y. To calculate the volume let x = 5 and y = 3.
This image cannot currently be displayed.
This image cannot currently be displayed.This image cannot currently be displayed.
xx
y
x2y = 52 · 3 = 25 · 3 = 75 cubic inches
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Multiplication of Polynomials
Multiplying MonomialsReview of the Distributive PropertiesMultiplying Monomials and PolynomialsMultiplying Polynomials
5.3
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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.
Slide 31Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 y4 3 26xx y y
1 4 3 26x y
5 56x y
Try some of Q: 7-16
Slide 32Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 33Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 25-32
Slide 34Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Multiplying Polynomials
Monomials, binomials, and trinomials are examples of polynomials.
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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
Try some of Q: 39-44
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Slide 37Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 45-64
Slide 38Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 39Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 65-72
Slide 40Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Multiplying polynomials vertically
Multiply 21 (2 3).x x x
22 3 1x x
x
22 3x x 3 22 3x x x 3 22 4 3x x x
Try some of Q: 73-78
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Special Products
Product of a Sum and DifferenceSquaring BinomialsCubing Binomials
5.4
Slide 42Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 43Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 44Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 27-32
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Slide 46Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 33-48
Slide 47Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 49-58
Slide 48Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Integer Exponents and the Quotient Rule
Negative Integers as ExponentsThe Quotient RuleOther Rules for ExponentsScientific Notation
5.5
Slide 50Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Negative Integers as Exponents
Simplify each expression.a. b. c.
Slide 51Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Evaluating negative exponents
Solution
a.
b.
c.
521
18
4( )a b
52
5
12
1
2 2 2 2 2
1
32
1
18
18 8
4( )a b 4
1( )a b
Try some of Q: 13-14
Evaluate the expression.
Slide 52Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Using the product rule with negative exponents
Solution
4 28 8
4 28 8 4 ( 2)8 28 64
Try some of Q: 15-18
Simplify the expression. Write the answer using positive exponents. a. b.
Slide 53Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
12y
Try some of Q: 25-36
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Simplify each expression. Write the answer using positive exponents.a. b. c.
Slide 55Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Using the quotient rule
Solution
a.
b.
c.
3
6
1010
7
3
xx
2 4
6
246
x yx y
3
6
1010
3 610 310 3
110
7 3x 4x
2 4
6
246
x yx y
2 4
6
246
x yx y
2 6 4 14x y
11000
7
3
xx
4 34x y3
4
4yx
Try some of Q: 36-40
Slide 56Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Simplify each expression. Write the answer using positive exponents.
a. b. c.
Slide 57Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Working with quotients and negative exponents
Solutiona.
b.
3
13
3 6
5 4
26
a ba b
33 27
c.
3 6
5 4
26
a ba b
4 6
5 3
26
b ba a
10
83ba
3
13
2
3
3a
2
3
3a
23
3a
6
23a
6
9a
Try some of Q: 41-48
Slide 58Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Important Powers of 10
Slide 59Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 60Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Try some of Q: 57-68
Slide 61Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Write each number in scientific notation.
a. 475,000 b. 0.00000325
475000
Slide 62Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Try some of Q: 69-80
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Division of Polynomials
Division by a MonomialDivision by a Polynomial
5.6
Slide 64Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Dividing a polynomial by a monomial
Divide.5 3
2
6 186
x xx
3
2
56 186
x xx
2 2
5 36 86 6
1x xx x
3 3x x
Try some of Q: 15-22
Slide 65Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Dividing and checking5 4 2
3
16 12 84
y y yy
Check:
5 4 2
3 3 3
16 12 84 4 4
y y yy y y
2 24 3y yy
3 2 24 4 3y y yy
3 2 3 3 24 4 4 3 4y y y y y
y
5 4 216 12 8y y y
Divide the expression and check the result.
5 4 2
3
16 12 84
y y yy
Try some of Q: 9-14
Slide 66Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 – 2x8x – 88x – 4
− 4
+ 4
42 4 .2 1
xx
RemainderQuotient + Divisor
Divide and check.24 6 82 1
x xx
Slide 67Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Try some of Q: 23-28
Slide 68Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 + 0x2x2 − 4x
4x − 8
+ 2x + 4
04x − 8
2 2 4.x x Try some of Q: 31-34
Slide 69Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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 − 2x2x3 + 0 − 4x
−5x2 + 2x + 5
+ 2x − 5
2x – 5 −5x2 + 0 + 10
22
2 53 2 5 .2
xx xx
Try some of Q: 35-38
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.