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College Algebra

On the cover:A colored version of the Flammarion Engraving. The black and white version of this engraving was included in the 1888 book by Camille Flammarion L’Atmosphère - Météorologie Populaire, which was a bookfor general audiences on meteorology. The original engraving was captioned: “Un missionaire du moyenâge raconte qu’il avait trouvé le point où le ciel et la Terre se touche.” This translates as “A missionaryof the Middle Ages told of how he found the point where the sky and the earth touch.”The Flammarion Engraving is often interpreted as representing the human quest to discover the innerworkings of the universe. This colored version is from the Wikimedia Commons and is credited to HugoHeikenwaelder.

C OLLEGE A LGEBRARichard W. BeveridgeSeptember 2, 2018

c 2018 Richard W. BeveridgeThis work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 UnportedLicense. The essence of the license is thatYou are free: to Share to copy, distribute and transmit the work to Remix to adapt the workUnder the following conditions: Attribution You must attribute the work in the manner specified by the author (but not inany way that suggests that they endorse you or your use of the work). Please contact the authorat rbeveridge@clatsopcc.edu to determine how best to make any attribution. Noncommercial You may not use this work for commercial purposes. Share Alike If you alter, transform, or build upon this work, you may distribute the resultingwork only under the same or similar license to this one.With the understanding that: Waiver Any of the above conditions can be waived if you get permission from the copyrightholder. Public Domain Where the work or any of its elements is in the public domain under applicablelaw, that status is in no way affected by the license. Other Rights In no way are any of the following rights affected by the license: Your fair dealing or fair use rights, or other applicable copyright exceptions and limitations; The author’s moral rights; Rights other persons may have either in the work itself or in how the work is used, such aspublicity or privacy rights. Notice For any reuse or distribution, you must make clear to others the license terms of thiswork. The best way to do this is with a link to the web page below.To view a full copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ or send a letterto Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.

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Contents12Algebra Review51.1Algebraic Simplification . . . . . . . . . . . . . . . . . . . . . . . . .51.2Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .101.3Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .251.4Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . .341.5Quadratic Equations with Complex Roots . . . . . . . . . . . . . . .431.6Multiplying and Dividing Rational Expressions . . . . . . . . . . .501.7Adding and Subtracting Rational Expressions . . . . . . . . . . . .581.8Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . .641.9Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .67Polynomial and Rational Functions832.1Representing Intervals . . . . . . . . . . . . . . . . . . . . . . . . . .832.2Solution by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . .882.3Solution of Polynomial Inequalities by Graphing . . . . . . . . . . .923

43CONTENTS2.4Solution of Rational Inequalities by Graphing . . . . . . . . . . . . . 1002.5Finding Factors from Roots . . . . . . . . . . . . . . . . . . . . . . . 1082.6Polynomial Long Division . . . . . . . . . . . . . . . . . . . . . . . . 1112.7Synthetic Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1202.8Roots and Factorization of Polynomials . . . . . . . . . . . . . . . . 128Exponents and Logarithms1353.1Exponential and Logistic Applications . . . . . . . . . . . . . . . . . 1353.2Logarithmic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1573.3Solving Exponential Equations . . . . . . . . . . . . . . . . . . . . . 1613.4Solving Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . 1683.5Applications of the Negative Exponential Function . . . . . . . . . 175Answer Key185

Chapter 1Algebra ReviewThis College Algebra text will cover a combination of classical algebra and analytic geometry, with an introduction to the transcendental exponential and logarithmic functions. If mathematics is the language of science, then algebra is thegrammar of that language. Like grammar, algebra provides a structure to mathematical notation, in addition to its uses in problem solving and its ability tochange the appearance of an expression without changing the value.1.1Algebraic SimplificationWhen algebraic techniques are presented as skills in isolation, they are much simpler to understand and practice. However the problem solving process in anycontext involves deciding which skills to use when. Most College Algebra students will have practiced problems in the form:(x 7)(x 2) ?or(2x 1)2 ?The problems in this section deal with a combination of these processes whichare often encountered as parts of more complex problems.5

6CHAPTER 1. ALGEBRA REVIEWExamplesSimplify:3(x 1)(2x 5) (x 4)2In this example, the simplification involves two expressions: 3(x 1)(2x 5) and(x 4)2 . The (x 4)2 is preceded by a negative (or subtraction) sign. This textbook will often treat x and ( x) as equivalent statements, since subtraction isdefined as the addition of a negative.We will simplify each expression separately and then look to combine like terms.3(x 1)(2x 5) (x 4)2 3(2x2 3x 5) (x2 8x 16)Notice that the results of both multiplications remain inside of parentheses. Thisis because each one has something that must be distributed.In the case of (2x2 3x 5), there is a 3 which must be distributed, resulting in6x2 9x 15. In the case of (x2 8x 16) there is a negative sign or 1 whichmust be distributed, resulting in x2 8x 16. It is important in these situationsthat the negative sign be distributed to all terms in the parentheses.So:3(x 1)(2x 5) (x 4)2 3(2x2 3x 5) (x2 8x 16) 6x2 9x 15 x2 8x 16 5x2 x 31Simplify:2(x 3)2 4(3x 1)(x 2)This example shows some of the same processes as the previous example. Thereare again two expressions that must be simplified, each of which has a coefficientthat must be distributed. It is often helpful to wait until after multiplying thebinomials before distributing the coefficient. However, as is often true in math-

1.1. ALGEBRAIC SIMPLIFICATION7ematics, there are several different approaches that may be taken in simplifyingthis problem.If someone prefers to first distribute the coefficient before multiplying the binomials, then the coefficient must only be distributed to ONE of the binomials, butnot both. For example, in multiplying 3 2 5 30, we can first multiply 2 5 10and then 3 10 30. Each factor is multiplied only once.In the example above we can proceed as we did with the previous example:2(x 3)2 4(3x 1)(x 2) 2(x2 6x 9) 4(3x2 5x 2) 2x2 12x 18 12x2 20x 8 10x2 8x 26Or, we can choose to distribute the 4 first:2(x 3)2 4(3x 1)(x 2) 2(x2 6x 9) (12x 4)(x 2) 2x2 12x 18 (12x2 20x 8) 2x2 12x 18 12x2 20x 8 10x2 8x 26Or, we can distribute the 4 as a negative. If we do this, then the sign in front ofthe parentheses will be positive:2(x 3)2 4(3x 1)(x 2) 2(x2 6x 9) ( 12x 4)(x 2) 2x2 12x 18 ( 12x2 20x 8) 10x2 8x 26Distributing the 2 in front of the squared binomial must also be handled carefullyif you choose to do this. If you distribute the 2 before squaring the (x 3), thenthe 2 will be squared as well. If you choose to distribute the 2, the (x 3)2 mustbe written out as (x 3)(x 3):

8CHAPTER 1. ALGEBRA REVIEW2(x 3)2 4(3x 1)(x 2) 2(x 3)(x 3) 4(3x 1)(x 2) (2x 6)(x 3) 4(3x2 5x 2) 2x2 12x 18 12x2 20x 8 10x2 8x 26Most examples in this text will distribute the coefficients as the last step beforecombining like terms for a final answer.Simplify:3x[5 (2x 7)] (3x 2)2 (x 5)(x 4)This example has three expressions that should be simplified separately beforecombining like terms. In the first expression 3x[5 (2x 7)], we should simplifyinside the brackets before distributing the 3x.3x[5 (2x 7)] (3x 2)2 (x 5)(x 4) 3x[5 2x 7] (3x 2)2 (x 5)(x 4) 3x[ 2x 2] (3x 2)(3x 2) (x2 x 20) 6x2 6x (9x2 12x 4) x2 x 20 6x2 6x 9x2 12x 4 x2 x 20 2x2 17x 24

1.1. ALGEBRAIC SIMPLIFICATIONExercises 1.1Simplify each expression.1)(x 2)[2x 2(3 x)] (x 5)22)3x2 [7x 2(2x 1)(3 x)]3)(a b)2 (a b)(a b) [a(2b 2) (b2 2a)]4)5x 3(x 2)(x 7) 3(x 2)25)(m 3)(m 1) (m 2)2 46)(a 1)(a 2) (a 2)(a 3) (a 3)(a 4)7)2a2 3(a 1)(a 2) [7 (a 1)]28)2(x 5)(3x 1) (2x 1)29)6y (3y 1)(y 2) (y 3)(y 8)10)6x 4(x 10)(x 1) (x 1)29

101.2CHAPTER 1. ALGEBRA REVIEWFactoringThis section will review three of the most common types of factoring - factoringout a Greatest Common Factor, Trinomial Factoring and factoring a Difference ofSquares.Greatest Common FactorFactoring out a greatest common factor essentially undoes the distributive multiplication that often occurs in mathematical expressions. This factor may be monomial or polynomial, but in these examples, we will explore monomial commonfactors.In multiplying 3xy 2 (5x 2y) 15x2 y 2 6xy 3 the monomial term 3xy 2 is multipliedor distributed to both terms inside the parentheses. The process of factorizationundoes this multiplication.Example:Factor 7x2 14xThis expression has two terms. The coefficients share a common factor of 7 andthe only variable involved in this expression is x. The highest power of the variable that is shared by both terms is x1 , so this is the power of x that can be factoredout of both terms. The greatest common factor is 7x.7x2 14x 7x(x 2)It isn’t necessary to find the greatest common factor right away. In more complicated problems, the factoring can be accomplished in pieces, similar in fashion toreducing fractions.

1.2. FACTORING11Example:Factor 42x2 y 6 98xy 3 210x3 y 2This expression has three terms. It’s not immediately clear what the greatestcommon factor of the coefficients is, but they’re all even numbers, so we could atleast divide them all by 2. The 98xy 3 term has an x1 , which means that this is thehightest power of x that we could factor out of all the terms. The 210x3 y 2 has ay 2 , which is the highest power of y that can be factored out of all the terms. So wecan at least proceed with these factors:42x2 y 6 98xy 3 210x3 y 2 2xy 2 21xy 4 2xy 2 49y 2xy 2 105x2 2xy 2 (21xy 4 49y 105x2 )Now, we didn’t try very hard to find the greatest common factor in the beginningof this problem, so it’s important that we continue to question whether or notthere are any remaining common factors. The 21 and 49 clearly share a commonfactor of 7, so it would make sense to see if 105 is divisible by 7 as well. If wedivide 105 by 7, we see that 105 7*15. So, we can also factor out a commonfactor of 7 from the remaining terms in the parentheses.2xy 2 (21xy 4 49y 105x2 ) 2xy 2 (7 3xy 4 7 7y 7 15x2 ) 7 2xy 2 (3xy 4 7y 15x2 ) 14xy 2 (3xy 4 7y 15x2 )

12CHAPTER 1. ALGEBRA REVIEWTrinomial Factoring (a 1)Trinomial factoring undoes the multiplication of two binomials, and it comes intwo flavors - simple and complex. The simplest form of trinomial factoring involves a trinomial expression in the form ax2 bx c in which the value of a is 1.This makes the task of factorization simpler than if the value of a is not 1.ExampleFactor x2 7x 10In this example, the value of a is 1, which makes this type of trinomial factoringa little less difficult that it would otherwise be. Whether or not the value of a is 1,the fundamental issue that governs this type of factoring is the or sign of theconstant term. In this problem, the constant term is positive. That means that weneed to find factors of 10 that add up to 7. This is relatively straightforward:x2 7x 10 (x 2)(x 5)A companion problem to this one is x2 7x 10. Notice that, in this case, thesign of the constant term is still positive, which means that we still need factorsof 10 that add up to 7. This means we still need to use 2 and 5. However, inthis case, instead of the 10 being produced from a multiplication of ( 2)( 5),it is the result of multiplying ( 2)( 5). This is what makes the 7 in the secondexample negative:x2 7x 10 (x 2)(x 5)ExampleFactor x2 3x 10In this case, the sign of the constant term is negative. That means that we need tofind factors of 10 that have a difference of 3. This is still 5 and 2.

1.2. FACTORING13x2 3x 10 (x 2)(x 5)The multiplication of the ( 2) and the ( 5) produce the ( 10) and the fact thatthe 2 and 5 have opposite signs creates the difference that gives us ( 3). A companion problem to this one is x2 3x 10. In this case, the sign of the constantterm is still negative, which means that we still need factors of 10 that have adifference of 3. This means we still need to use 2 and 5. However in this case,instead of the ( 3) as the coefficient of the middle term, we’ll need a ( 3). To dothis we simply reverse the signs of the 2 and 5 from the previous problem:x2 3x 10 (x 2)(x 5)Now the ( 2)( 5) gives us ( 10), but the 2x 5x gives us ( 3x) instead of( 3x).ExampleFactor x2 11x 42In this problem, the sign of the constant ter

On the cover: A colored version of the Flammarion Engraving. The black and white version of this engraving was in-cluded in the 1888 book by Camille Flammarion L’Atmosph ere - M