Math 120 Textbook.pdf - In nterm mediate Algeb Textbbookk A...

This preview shows page 1 out of 525 pages.

Unformatted text preview: In nterm mediate Algeb Textbbookk A bra T for Skyli S ine C Colleege Adapted for Skyline College from text created by Department of Mathematics College of the Redwoods Acknowledgments Authored and edited by David Arnold, Michael Butler, Mike Haley, Diane Harrow, Aeron Ives, Stephen Jackson, Craig Kutil, Teresa Matsumoto, Jacob Miles Prystowsky, Todd Olsen, David Tuttle, and Bruce Wagner. Design, layout, and technical work by Jacob Miles Prystowsky. Additional assitance from Aditya Mahajan, Hans Hagen, Taco Hoekwater, and the ConTEXt mailing list. Funded by a grant from the Project for Learning Enhancement. Skyline College Edition Contributors Department of Mathematics, Skyline College Cover photo by Tadashi Tsuchida Changes to the Skyline College Edition: Chapter 2.5 Chapter 4.1 Copyright All parts of this intermediate algebra textbook are copyrighted in the name of Department of Mathematics, College of the Redwoods. They are not in the public domain. However, they are being made available free for use in educational institutions. This offer does not extend to any application that is made for profit. Users who have such applications in mind should contact David Arnold at [email protected] or Bruce Wagner at [email protected] This work (including all text, Portable Document Format files, and any other orig-inal works), except where otherwise noted, is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 2.5 License, and is copyrighted ©2006, Department of Mathematics, College of the Redwoods. To view a copy of this license, visit or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA. Contents Chapter 1 Factoring ................................................................................................................... 2 1.1 The Greatest Common Factor ..................................................................................... 2 1.2 Solving Nonlinear Equations...................................................................................... 15 1.3 Factoring ax2+bx+c where a=1.................................................................................. 29 1.4 Factoring ax2+bx+c where a1 .................................................................................. 43 1.5 Factoring Special Forms ............................................................................................ 56 1.6 Applications of Factoring ............................................................................................ 71 Chapter 2 Quadratic Functions ............................................................................................ 79 2.1 The Parabola ................................................................................................................ 80 2.2 Vertex Form ................................................................................................................ 101 2.3 Zeros of the Quadratic .............................................................................................. 119 2.4 The Quadratic Formula ............................................................................................. 139 2.5 Complex Numbers ..................................................................................................... 161 2.6 Optimization................................................................................................................ 167 Chapter 3 Rational Functions ............................................................................................. 183 3.1 Introducing Rational Functions ................................................................................ 184 3.2 Reducing Rational Functions ................................................................................... 199 3.3 Products and Quotients of Rational Functions ..................................................... 216 3.4 Sums and Differences of Rational Functions ........................................................ 233 3.5 Complex Fractions..................................................................................................... 247 3.6 Solving Rational Equations ...................................................................................... 262 3.7 Applications of Rational Functions.......................................................................... 279 3.8 Direct and Inverse Variation .................................................................................... 295 Chapter 4 Exponential and Logarithmic Functions ...................................................... 304 4.1 Introduction to Exponential Functions .................................................................... 305 4.2 Exponents and Roots................................................................................................ 317 4.3 Exponential Functions .............................................................................................. 336 4.4 Applications of Exponential Functions ................................................................... 353 4.5 Inverse Functions ...................................................................................................... 367 4.6 Logarithmic Functions ............................................................................................... 385 4.7 Properties of Logarithms .......................................................................................... 395 4.8 Exponential Growth and Decay ............................................................................... 406 Chapter 5 Radical Functions............................................................................................... 419 5.1. The Square Root Function ....................................................................................... 420 5.2 Multiplication Properties or Radicals ...................................................................... 435 5.3 Division Properties of Radicals ................................................................................ 455 5.4 Radical Expressions .................................................................................................. 472 5.5 Radical Equations ...................................................................................................... 491 5.6 The Pythagorean Theorem ...................................................................................... 505 Index .......................................................................................................................................... 526 $IBQUFS  FBDUPSJOH   5IF BODJFOU #BCZMPOJBOT MFGU UIF FBSMJFTU FWJEFODF PG UIF VTF PG RVBESBUJD FRVBUJPOT PO DMBZ UBCMFUTEBUJOHCBDLUP#$5IFZVOEFSTUPPEIPXUIFBSFBPGBTRVBSFDIBOHFTXJUIUIF MFOHUIPGJUTTJEF'PSFYBNQMF UIFZLOFXJUXBTQPTTJCMFUPTUPSFOJOFUJNFTNPSFCBMFTPGIBZ JOBTRVBSFMPGUJGUIFTJEFPGUIFMPGUXBTUSJQMFEJOMFOHUI)PXFWFS UIFZEJEOPULOPXIPXUP DBMDVMBUFUIFMFOHUIPGUIFTJEFPGBTRVBSFTUBSUJOHGSPNBHJWFOBSFB5IFXPSEiRVBESBUJDw DPNFT GSPN iRVBESBUVT w UIF -BUJO XPSE GPS iTRVBSFw *O UIJT DIBQUFS  XF XJMM MFBSO IPX UP TPMWFDFSUBJORVBESBUJDFRVBUJPOTCZGBDUPSJOHQPMZOPNJBMT 1 1 The (SFBUFTU$PNNPO'BDUPS We begin this section with definitions of factors and divisors. Because 24 = 2 · 12, both 2 and 12 are factors of 24. However, note that 2 is also a divisor of 24, because when you divide 24 by 2 you get 12, with a remainder of zero. Similarly, 12 is also a divisor of 24, because when you divide 24 by 12 you get 2, with a remainder of zero. Factors and divisors. Suppose m and n are integers. Then m is a divisor (factor ) of n if and only if there exists another integer k so that n = m · k. The words divisor and factor are equivalent. They have the same meaning. You Try It! List the positive divisors of 18. EXAMPLE 1. List the positive divisors (factors) of 24. Solution: First, list all possible ways that we can express 24 as a product of two positive integers: 24 = 1 · 24 Answer: 1, 2, 3, 6, 9, and 18 or 24 = 2 · 12 or 24 = 3 · 8 or 24 = 4 · 6 Therefore, the positive divisors (factors) of 24 are 1, 2, 3, 4, 6, 8, and 24.  You Try It! List the positive divisors that 40 and 60 have in common. EXAMPLE 2. common. List the positive divisors (factors) that 36 and 48 have in Solution: First, list all positive divisors (factors) of 36 and 48 separately, then box the divisors that are in common. Divisors of 36 are: 1 , 2 , 3 , 4 , 6 , 9, 12 , 18, 36 Divisors of 48 are: 1 , 2 , 3 , 4 , 6 , 8, 12 , 16, 24, 48 Answer: 1, 2, 4, 5, 10, and 20 Therefore, the common positive divisors (factors) of 36 and 48 are 1, 2, 3, 4, 6, and 12.  2 5IF(SFBUFTU$PNNPO'BDUPS Greatest common divisor. The greatest common divisor (factor) of a and b is the largest positive number that divides evenly (no remainder) both a and b. The greatest common divisor of a and b is denoted by the symbolism GCD(a, b). We will also use the abbreviation GCF(a, b) to represents the greatest common factor of a and b. Remember, greatest common divisor and greatest common factor have the same meaning. In Example 2, we listed the common positive divisors of 36 and 48. The largest of these common divisors was 12. Hence, the greatest common divisor (factor) of 36 and 48 is 12, written GCD(36, 48) = 12. With smaller numbers, it is usually easy to identify the greatest common divisor (factor). You Try It! EXAMPLE 3. State the greatest common divisor (factor) of each of the following pairs of numbers: (a) 18 and 24, (b) 30 and 40, and (c) 16 and 24. State the greatest common divisor of 36 and 60. Solution: In each case, we must find the largest possible positive integer that divides evenly into both the given numbers. a) The largest positive integer that divides evenly into both 18 and 24 is 6. Thus, GCD(18, 24) = 6. b) The largest positive integer that divides evenly into both 30 and 40 is 10. Thus, GCD(30, 40) = 10. c) The largest positive integer that divides evenly into both 16 and 24 is 8. Thus, GCD(16, 24) = 8. Answer: 12  With larger numbers, it is harder to identify the greatest common divisor (factor). However, prime factorization will save the day! You Try It! EXAMPLE 4. Find the greatest common divisor (factor) of 360 and 756. Solution: Prime factor 360 and 756, writing your answer in exponential form. 3 Find the greatest common divisor of 120 and 450. $IBQUFS'BDUPSJOH 360 756 10 36 4 2 9 2 3 84 9 2 5 3 3 3 7 12 3 4 2 2 Thus: 360 = 23 · 32 · 5 756 = 22 · 33 · 7 To find the greatest common divisor (factor), list each factor that appears in common to the highest power that appears in common. In this case, the factors 2 and 3 appear in common, with 22 being the highest power of 2 and 32 being the highest power of 3 that appear in common. Therefore, the greatest common divisor of 360 and 756 is: GCD(360, 756) = 22 · 32 =4·9 = 36 Therefore, the greatest common divisor (factor) is GCD(360, 756) = 36. Note what happens when we write each of the given numbers as a product of the greatest common factor and a second factor: 360 = 36 · 10 756 = 36 · 21 Answer: 30 In each case, note how the second second factors (10 and 21) contain no additional common factors.  Finding the Greatest Common Factor of Monomials Example 4 reveals the technique used to find the greatest common factor of two or more monomials. Finding the GCF of two or more monomials. To find the greatest common factor of two or more monomials, proceed as follows: 4 5IF(SFBUFTU$PNNPO'BDUPS 1. Find the greatest common factor (divisor) of the coefficients of the given monomials. Use prime factorization if necessary. 2. List each variable that appears in common in the given monomials. 3. Raise each variable that appears in common to the highest power that appears in common among the given monomials. You Try It! EXAMPLE 5. Find the greatest common factor of 6x3 y 3 and 9x2 y 5 . Solution: To find the GCF of 6x3 y 3 and 9x2 y 5 , we note that: Find the greatest common factor of 16xy 3 and 12x4 y 2 . 1. The greatest common factor (divisor) of 6 and 9 is 3. 2. The monomials 6x3 y 3 and 9x2 y 5 have the variables x and y in common. 3. The highest power of x in common is x2 . The highest power of y in common is y 3 . Thus, the greatest common factor is GCF(6x3 y 3 , 9x2 y 5 ) = 3x2 y 3 . Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial: 6x3 y 3 = 3x2 y 3 · 2x 9x2 y 5 = 3x2 y 3 · 3y Observe that the set of second monomial factors (2x and 3y) contain no additional common factors. Answer: 4xy 2  You Try It! EXAMPLE 6. Find the greatest common factor of 12x4 , 18x3 , and 30x2 . Solution: To find the GCF of 12x4 , 18x3 , and 30x2 , we note that: 1. The greatest common factor (divisor) of 12, 18, and 30 is 6. 2. The monomials 12x4 , 18x3 , and 30x2 have the variable x in common. 3. The highest power of x in common is x2 . 5 Find the greatest common factor of 6y 3 , 15y 2 , and 9y 5 . $IBQUFS'BDUPSJOH Thus, the greatest common factor is GCF(12x4 , 18x3 , 30x2 ) = 6x2 . Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial: 12x4 = 6x2 · 2x2 18x3 = 6x2 · 3x 30x2 = 6x2 · 5 Answer: 3y 2 Observe that the set of second monomial factors (2x2 , 3x, and 5) contain no additional common factors.  Factor Out the GCF In Chapter 5, we multiplied a monomial and polynomial by distributing the monomial times each term in the polynomial. 2x(3x2 + 4x − 7) = 2x · 3x2 + 2x · 4x − 2x · 7 = 6x3 + 8x2 − 14x In this section we reverse that multiplication process. We present you with the final product and ask you to bring back the original multiplication problem. In the case 6x3 + 8x2 − 14x, the greatest common factor of 6x3 , 8x2 , and 14x is 2x. We then use the distributive property to factor out 2x from each term of the polynomial. 6x3 + 8x2 − 14x = 2x · 3x2 + 2x · 4x − 2x · 7 = 2x(3x2 + 4x − 7) Factoring. Factoring is “unmultiplying.” You are given the product, then asked to find the original multiplication problem. First rule of factoring. If the terms of the given polynomial have a greatest common factor (GCF), then factor out the GCF. Let’s look at a few examples that factor out the GCF. You Try It! Factor: 9y 2 − 15y + 12 EXAMPLE 7. Factor: 6x2 + 10x + 14 6 5IF(SFBUFTU$PNNPO'BDUPS Solution: The greatest common factor (GCF) of 6x2 , 10x and 14 is 2. Factor out the GCF. 6x2 + 10x + 14 = 2 · 3x2 + 2 · 5x + 2 · 7 = 2(3x2 + 5x + 7) Checking your work. Every time you factor a polynomial, remultiply to check your work. Check: Multiply. Distribute the 2. 2(3x2 + 5x + 7) = 2 · 3x2 + 2 · 5x + 2 · 7 = 6x2 + 10x + 14 Answer: 3(3y 2 − 5y + 4) That’s the original polynomial, so we factored correctly.  You Try It! EXAMPLE 8. Factor: 12y 5 − 32y 4 + 8y 2 Factor: 8x6 + 20x4 − 24x3 Solution: The greatest common factor (GCF) of 12y 5 , 32y 4 and 8y 2 is 4y 2 . Factor out the GCF. 12y 5 − 32y 4 + 8y 2 = 4y 2 · 3y 3 − 4y 2 · 8y 2 + 4y 2 · 2 = 4y 2 (3y 3 − 8y 2 + 2) Check: Multiply. Distribute the monomial 4y 2 . 4y 2 (3y 3 − 8y 2 + 2) = 4y 2 · 3y 3 − 4y 2 · 8y 2 + 4y 2 · 2 = 12y 5 − 32y 4 + 8y 2 Answer: 4x3 (2x3 + 5x − 6) That’s the original polynomial. We have factored correctly.  You Try It! EXAMPLE 9. Factor: 12a3 b + 24a2 b2 + 12ab3 3 2 2 3 Solution: The greatest common factor (GCF) of 12a b, 24a b and 12ab is 12ab. Factor out the GCF. 12a3 b + 24a2 b2 + 12ab3 = 12ab · a2 − 12ab · 2ab + 12ab · b2 = 12ab(a2 + 2ab + b2 ) 7 Factor: 15s2 t4 + 6s3 t2 + 9s2 t2 $IBQUFS'BDUPSJOH Check: Multiply. Distribute the monomial 12ab. 12ab(a2 + 2ab + b2 ) = 12ab · a2 − 12ab · 2ab + 12ab · b2 = 12a3 b + 24a2 b2 + 12ab3 Answer: 3s2 t2 (5t2 + 2s + 3) That’s the original polynomial. We have factored correctly.  Speeding Things Up a Bit Eventually, after showing your work on a number of examples such as those in Examples 7, 8, and 9, you’ll need to learn how to perform the process mentally. You Try It! Factor: 18p5 q 4 − 30p4 q 5 + 42p3 q 6 EXAMPLE 10. Factor each of the following polynomials: (a) 24x + 32, (b) 5x3 − 10x2 − 10x, and (c) 2x4 y + 2x3 y 2 − 6x2 y 3 . Solution: In each case, factor out the greatest common factor (GCF): a) The GCF of 24x and 32 is 8. Thus, 24x + 32 = 8(3x + 4) b) The GCF of 5x3 , 10x2 , and 10x is 5x. Thus: 5x3 − 10x2 − 10x = 5x(x2 − 2x − 2) c) The GCF of 2x4 y, 2x3 y 2 , and 6x2 y 3 is 2x2 y. Thus: 2x4 y + 2x3 y 2 − 6x2 y 3 = 2x2 y(x2 + xy − 3y 2 ) As you speed things up by mentally factoring out the GCF, it is even more important that you check your results. The check can also be done mentally. For example, in checking the third result, mentally distribute 2x2 y times each term of x2 + xy − 3y 2 . Multiplying 2x2 y times the first term x2 produces 2x4 y, the first term in the original polynomial. 2x2 y (x2 + xy − 3y 2 ) = 2x4 y + 2x3 y 2 − 6x2 y 3 Answer: 6p3 q 4 (3p2 − 5pq + 7q 2 ) Continue in this manner, mentally checking the product of 2x2 y with each term of x2 + xy − 3y 2 , making sure that each result agrees with the corresponding term of the original polynomial.  Remember that the distributive property allows us to pull the GCF out in front of the expression or to pull it out in back. In symbols: ab + ac = a(b + c) 8 or ba + ca = (b + c)a 5IF(SFBUFTU$PNNPO'BDUPS You Try It! EXAMPLE 11. Factor: 2x(3x + 2) + 5(3x + 2) Solution: In this case, the greatest common factor (GCF) is 3x + 2. Factor: 3x2 (4x − 7) + 8(4x − 7) 2x(3x + 2) + 5(3x + 2) = 2x · (3x + 2) + 5 · (3x + 2) = (2x + 5)(3x + 2) Because of the commutative property of multiplication, it is equally valid to pull the GCF out in front. 2x(3x + 2) + 5(3x + 2) = (3x + 2) · 2x + (3x + 2) · 5 = (3x + 2)(2x + 5) Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same. Answer: (3x2 + 8)(4x − 7)  You Try It! EXAMPLE 12. Factor: 15a(a + b) − 12(a + b) Solution: In this case, the greatest common factor (GCF) is 3(a + b). Factor: 24m(m − 2n) + 20(m − 2n) 15a(a + b) − 12(a + b) = 3(a + b) · 5a − 3(a + b) · 4 = 3(a + b)(5a − 4) Alternate solution: It is possible that you might fail to notice that 15 and 12 are divisible by 3, factoring out only the common factor a + b. 15a(a + b) − 12(a + b) = 15a · (a + b) − 12 · (a + b) = (15a − 12)(a + b) However, you now need to notice that you can continue, factoring out 3 from both 15a and 12. = 3(5a − 4)(a + b) Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same. Answer: 4(6m + 5)(m − 2n)  9 $IBQUFS'BDUPSJOH Factoring by Grouping The final factoring skill in this section involves four-term expressions. The technique for factoring a four-term expression is called factoring by grouping. You Try It! Factor by grouping: x2 − 6x + 2x − 12 EXAMPLE 13. Factor by grouping: x2 + 8x + 3x + 24 Solution: We “group” the first and second terms, noting that we can factor an x out of both of these terms. Then we “group” the third and fourth terms, noting that we can factor 3 out of both of these terms. x2 + 8x + 3x + 24 = x (x + 8) + 3 (x + 8) Now we can factor x + 8 out of both of these terms. = (x + 3)(x + 8) Answer: (x + 2)(x − 6)  Let’s try a grouping that contains some negative signs. You Try It! Factor by grouping: x2 − 5x − 4x + 20 EXAMPLE 14. Factor by grouping: x2 + 4x − 7x − 28 Solution: We “group” the first and second terms, noting that we can factor x out of both of these terms. Then we “group” the third and fourth terms, then try to factor a 7 out of both these terms. x2 + 4x − 7x − 28 = x (x + 4) + 7 (−x − 4) This does not lead to a common factor. Let’s try again, this time factoring a −7 out of the third and fourth terms. x2 + 4x − 7x − 28 = ...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture